THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

LOS  ANGELES 

GIFT  OB 

John  S.Prell 


KEY 


TO 


WITH 


MANY  ADDITIONAL    EXAMPLES,  ILLUSTRATING 
THE    ALGEBKAIC    ANALYSIS. 


BY  CHAKLES  DAVIES,  LL.D., 

AUTHOR  OF  A  FULL  COURSE  OF  MATHEMATICS. 


S.  PRELL 

Lwil  &  Mechanical  Engineer 

SAN  THAN  CISCO,  CAL. 
A.   S.   BARNES    &   COMPANY, 

HEW  TOEK,  CHICAGO  AND  NEW  ORLEANS, 
1876. 


DAYIES'  COURSE  OF  MATHEMATICS, 


I3ST      THREE      ~E*  A.  E,  T  S  - 


I.     COMMON    SCHOOL    COURSE. 

DAVTES'   PRIMARY   ARITHMETIC.  — The  fundamental   principles   displayed   in   Object 
Lessons. 

DAVIES'  INTELLECTUAL  ARITHMETIC.— Referring  all  operations  to  the  unit  1  as  the  only 
tangible  basis  for  logical  development. 

DAVIES'  ELEMENTS  OP  WRITTEN  ARITHMETIC. — A  practical  introduction  to  the  whole 
subject.    Theory  subordinated  to  Practice. 

DAVIES'  PRACTICAL  ARITHMETIC,— A  combination  of  Theory  and  Practice,  clear,  exact, 
brief,  and  comprehensive. 

II.     ACADEMIC    COURSE. 

DAVIES'  UNIVERSITT  ARITHMETIC.— Treating  the  subject  exhaustively  as  a  science,  in  a 
logical  series  of  connected  propositions. 

DAVTES'  ELEMENTARY  ALGEBRA.— A  connecting  link,  conducting  the  pupil  easily  from 
arithmetical  processes  to  abstract  analysis. 

DAVIES'  UNIVERSITY  ALGEBRA. — For  institutions  desiring  a  more  complete  but  not  the 
fullest  course  in  pure  Algebra. 

DAVIES'  PRACTICAL  MATHEMATICS.— The  science  practically  applied  to  the  useful  art?, 
as  Drawing,  Architecture,  Surveying,  Mechanics,  etc. 

DAVIES'  ELEMENTARY  GEOMETRY.— The  important  principles  in  simple  form,  but  with 
all  the  exactness  of  rigorous  reasoning. 

DAVTES'  ELEMENTS  OP  SURVEYING.— Re-written  in  1870.   A  simple  and  full  presentation 
for  Instruction  and  Practice. 

III.     COLLEGIATE    COURSE. 

DAVIES'  BOURDON'S  ALGEBRA.— Embracing  Sturm's  Theorem,  and  a  most  exhaustive 
course.    Re-written,  in  1873. 

DAVIES'  UNIVERSITY  ALGEBRA. — A  shorter  coarse  than  Bourdon,  for  Institutions  hav- 
ing less  time  to  give  the  subject. 

DAVTES'  LEGENDRE'S  GEOMETRY.— A  standard  work  in  this  country  and  in  Europe. 

DAVIES'  ANALYTICAL  GEOMETRY.— A  full  course  of  Analysis,  embracing  the  applications 
to  surfaces  of  the  second  order. 

DAVTES'  DIFFERENTIAL  AND  INTEGRAL  CALCULUS,  on  the  basis  of  Continuous  Quantity 
and  Consecutive  Differences. 

DAVIES'  ANALYTICAL  GEOMETRY  AND  CALCULUS.— The  shorter  treatises,  combined  in 
one  volume. 

DAVTES'  DESCRIPTIVE  GEOMETRY.— With  application  to  Spherical  Trigonometry,  Spher- 
ical Projections,  and  Warped  Surfaces. 

DAVTES'  SHADES,  SHADOWS,  AND  PERSPECTIVE.— A  succinct  exposition  of  the  mathe- 
matical principles  involved. 

DAVTES'  NATURE  AND  UTILITY  OF  MATHEMATICS,  Logically  considered. 

DAVTES  AND  PECK'S  MATHEMATICAL  DICTIONARY,  or  Cyclopedia  of  Mathematics. 


Entered  according  to  Act  of  Congress,  in  the  year  1873,  by 

CHARLES    DAVIES, 
In  the  Office  of  the  Librarian  of  Congress,  at  Washington. 


PREFACE. 


A  WIDE  difference  of  opinion  is  known  to  exist  among  teachers 
in  regard  to  the  value  of  a  Key  to  any  mathematical  work,  and 
it  is  perhaps  yet  undecided  whether  a  Key  is  a  help  or  a 
hindrance. 

If  a  Key  is  designed  to  supersede  the  necessity  of  investiga- 
tion and  labor  on  the  part  of  the  teacher;  to  present  to  his 
mind  every  combination  of  thought  which  ought  to  be  suggested 
by  a  problem,  and  to  permit  him  to  float  sluggishly  along  the 
current  of  ideas  developed  by  the  author,  it  would  certainly  do 
great  harm,  and  should  be  excluded  from  every  school. 

If,  on  the  contrary,  a  Key  is  so  constructed  as  to  suggest 
ideas,  both  in  regard  to  particular  questions  and  general  science, 
which  the  Text-book  might  not  impart;  if  it  develops  jpnethods 
of  solution  too  particular  or  too  elaborate  to  find  a  place  in  the 
text ;  if  it  is  mainly  designed  to  lessen  the  mechanical  labor  of 
teaching,  rather  than  the  labor  of  study  and  investigation ;  it 
may,  in  the  hands  of  a  good  teacher,  prove  a  valuable  auxiliary. 

The    KEY   TO  BOURDON   is  intended   to  answer,    precisely,    this 

733414 


rr  PREFACE. 

end.  The  principles  developed  in  the  text  are  explained  and 
illustrated  by  means  of  numerous  examples,  and  these  are  all 
wrought  in  the  Key  by  methods  which  accord  with  and  make 
evident  the  principles  themselves.  The  Key,  therefore,  not  only 
explains  the  various  questions,  but  is  a  commentary  on  the  text 
itself. 

Nothing  is  more  gratifying  to  an  ambitious  teacher  than  to 
push  forward  the  investigations  of  his  pupils  beyond  the  limits 
of  the  text  book.  To  aid  him  in  an  undertaking  so  useful  to 
himself  and  to  them,  an  Appendix  has  been  added,  containing  a 
copious  collection  of  Practical  Examples.  Many  of  the  solutions 
are  quite  curious  and  instructive ;  and  taken  in  connection  with 
those  embraced  in  the  Text,  form  a  full  and  complete  system  of 
Algebraic  Analysis. 

The  many  letters  which.  I  have  received  from  Teachers  and 
Pupils,  in  regard  to  the  best  solutions  of  new  questions,  have 
suggested  the  desirableness  of  furnishing,  in  the  present  work, 
those  which  have  been  most  approved.  They  are  a  collection 
of  problems  that  have  special  values,  and  their  solutions  may  be 
studied  with  great  profit  by  every  one  seeking  mathematical 
knowledge. 

FISHKILL  LANDING, 


July 


,  LANDING,) 
',  1873.         f 


INTRODUCTION, 


ALGEBRA. 

1.    ON  an   analysis  of  the  subject  of  Algebra,  we     Aig*b». 
think  it  will  appear  that  the  subject  itself  presents  no     Difficult!.* 
serious  difficulties,  and  that  most  of  the  embarrassment 
which  is  experienced  by  the  pupil  in  gaining  a  knowl- 
edge of  its  principles,  as  well  as  in  their  applications, 
arises  from   not  attending   sufficiently  to  the   language     Langum*. 
or  signs  of  the  thoughts  which    are   combined  in  the 
reasonings.     At  the  hazard,  therefore,  of  being  a  little 
diffuse,    I   shall   begin  with    the  very  elements  of  the 
algebraic   language,   and   explain,   with    much    minute- 
ness, the  exact  signification  of  the  characters  that  stand     Charge™ 

•which  repre- 

for  the  quantities  which  are  the  subjects  of  the  analy-  sent  quantity 
sis  ;  and  also  of  those  signs  which  indicate  the  several        Sign*. 
operations  to  be  performed  on  the  quantities. 


2.   The   quantities   which   are    the   subjects   of  the     n 
algebraic   analysis   may   be   divided    into  two  classes:    HowdiTid«A 
those  which  are  known  or  given,  and  those  which  are 
unknown  or  sought.     The  known  are  uniformily  repre-     ^ 
sented  by  the  first  letters  of  the  alphabet,  a,  b,  c,  d, 
&c.  ;   and  the   unknown  by  the  final  letters,  a-,  y,  z, 
v,  w,  &c. 


6  INTRODUCTION. 

^rLseVor         Quantity  is   susceptible  of  being  increased,  di- 
diminished.    minisbed,  and  measured  ;  and  there  are  six  operations 
Five  opera-     which  can  be  performed  upon  a  quantity  that  will 
give  results  differing  from  the  quantity  itself,  viz.: 
First  1st.  To  add  it  to  itself  or  to  some  other  quantity ; 

second  2d.    To  subtract  some  other  quantity  from  it; 

Third.  3d.    To  multiply  it  by  a  number; 

4th.  To  divide  it; 

Fourth. 

5th.  To  raise  it  to  any  power; 

Fifth. 

6th.  To  extract  a  root  of  it. 

The  cases  in  which  the  multiplier  or  divisor  is  1, 
are  of  course  excepted;  as  also  the  case  where  a 
root  is  to  be  extracted  of  1. 

signs.  4  The  six  signs  which  denote  these  operations 
Elements  are  ^OQ  wejj  known  to  be  repeated  here.  These,  with 
Algebraic  tne  gigns  of  equality  and  inequality,  the  letters  of  the 

language. 

alphabet  and  the  figures  which  are  employed,  make  up 
lu  words  and  fae    elements  of   the  algebraic    language.     The    words 

phrases  : 

and    phrases    of    the    algebraic,    like    those    of    every 
HO.W  inter-     other   language,  are    to   be    taken    in    connection  with 

preted. 

each  other,  and  are  not   to  be   interpreted  as  separate 
and  isolated  symbols. 

5.  The    symbols  of  quantity  are    designed  to  repre- 
sent quantity  in  general,  whether  abstract  or  concrete, 
General.       whether  known  or    unknown ;  and  the  signs  which  in- 
dicate the    operations  to  be    performed  on  the  quanti- 
Exampies.     ties  are  to  be  interpreted   in  a  sense    equally  general. 
When    the    sign   plus   is  written,  it  indicates  that  the 
SiB  minus.™*  quality  before    which    it   is  placed  is  to  be  added  to 
some  other    quantity :  and   the  sign  minus   implies  the 


INTRODUCTION.  7 

existence  of  a  minuend,  from  which  the  subtrahend  is 

to  be  taken.     One  thing  should  be  observed  in  regard  s'£ns  h»T«  •« 

effect  on  the 

to  the  signs  which  indicate  the  operations  that  are  to       nature  of 

be   performed   on   quantities,  viz. :    they  do    not   at   all 

affect  or  change  the   nature  of  the  quantity  before   or 

after  which   they  are  written,  but  merely  indicate  what 

is  to  be  done  with  t//e  quantify.     In  Algebra,  for  ex-     Example*: 

In  Algebr*. 

ample,  the  minus  sign  merely  indicates  that  the  quan- 
tity before  which  it  is  written  is  to  be  subtracted  from 
some  other  quantity ;  and  in  Analytical  Geometry,  that  in  Analytical 
the  line  before  which  it  falls  is  estimated  in  a  contrary 
direction  from  that  in  which  it  would  have  been  reck- 
oned, had  it  had  the  sign  plus  ;  but  in  neither  case  is 
the  nature  of  the  quantity  itself  different  from  what 
it  would  have  been  had  the  sign  been  plus. 

The   interpretation    of   the   language   of  Algebra   is  interpretation 

of  the 

the  first  thing  to  which  the  attention  of  a  pupil  should      language-. 

be  directed ;  and  he  should  be  drilled  on  the  meaning 

and   import   of  the   symbols,   until    their    significations 

and  uses  are   as  familiar  as  the  sounds  and  combina-    *••*•••*•* 

tions  of  the  letters  of  the  alphabet. 

6.   Beginning   with    the  elements   of  the   language,      Elements 

explained 

let  any  number  or  quantity  be  designated  by  the  letter 
a,  and  let  it  be  required  to  add  this  letter  to  itself 
and  find  the  result  or  sum.  The  addition  will  be 

expressed  by 

a  -f  a  =  the  sum. 

But  how   is  the   sum   to   be   expressed?     By   simply    Signification 
regarding  a  as  one  a,  or  la,  and  then  observing  that 
one  a  and  one  a,  make  tico  a's    or  2a :  hence, 


INTRODUCTION. 

a  +  a  =  2a  ; 


and  thus  we  place  a  figure  before  a  letter  to  indicate 
how  many  times  it  is  taken.     Such  figure  is  called  a 

Co-efEcient 


p»anct:  7.    The   product   of    several    numbers   is   indicated 

by  the  sign  of  multiplication,  or  by  simply  writing  the 
letters  which  represent  the  numbers  by  the  side  of 
each  other.  Thus, 

4ow  indicated  ttXbXCXdxf,     OT     abcdf, 

indicates  the  continued  product  of  a,  i,  c,  d,  and  /, 
r»ctor.  and  each  letter  is  called  a  factor  of  the  product  : 
hence,  a  factor  of  a  product  is  one  of  the  multipliers 
which  produce  it.  Any  figure,  as  5,  written  before  a 
product,  as 

5abcdf, 

Co-efficient  of  is  the  co-efficient  of  the  product,  and  shows  that  the 
product  is  taken  5  times. 


actors  :        g^    jf   tjje   numbers  represented   by  a,  b,  c,  d,  and 
wh»t  the      _/,   were   equal    to   each    other,   they   would    each    be 

product 

become*.      represented    by   a    single    letter   a,   and    the    product 
would   then   become 

How 


that  is,  we  indicate  the  product  of  several  equal  fac- 
tors by  simply  writing  the  letter  once  and  placing  a 
figure  above  and  a  Itole  at  the  right  >f  it,  to  indicate 


INTRODUCTION.  9 

how  many  times  it  is  taken  as  a  factor.     The  figure     Exponent: 
BO  written  is  called  an   exponent.     Hence,  an  exponent  where  writua. 
is  a  simple  form  of   language    to  point  out  how  many 
equal  factors  are  employed. 

9.  The  division  of  one  quantity  by  another  is  indi-     Diri«ion: 
cated  by  simply  writing  the  divisor  below  the  dividend,         hoir 

n  />  expreised 

after  the  manner  ot  a  traction  ;  by  placing  it  on  the 
right  of  the  dividend  with  a  horizontal  line  and  two 
dots  between  them ;  or  by  placing  it  on  the  right  with 
a  vertical  line  between  them :  thus  either  form  of 
expression : 

a         . 
— ,  b  —  a,     or    6  |  a,  Three  fonnit 

indicates  the  division  of  b  by  a. 

10.  The   extraction  of  a  root   is   indicated  by  the        Roots : 
sign  -y/.     This  sign,  when  used  by  itself  indicates  the  howindicaud 
lowest  root,  viz.,  the  square  root.      If  any  other  root 

is  to  be  extracted,  as  the  third,  fourth,  fifth,  &c.,  the        index; 
figure  marking  the  degree  of  the  root  is  written  above  where  writt«m 
and  at  the  left  of  the  sign ;  as, 

^/~cube  root,  ^/""fourth  root,  &c. 

The  figure  so  written,  is  called  the  Index  of  the  root.    Language  fa 

We  have  thus   given   the  very  simple  and   general     0p«r«titL 
language    by    which    we    indicate    each    of  the    five 
operations    that    may   be  performed   on    an    algebraic 
quantity,  and  every  process  in  Algebra  involves  one  or 
other  of  these  operations. 


10  INTRODUCTION. 

MINUS     SIGN. 

language":  U-    The   algebraic   symbols    are   divided    into   two 

classes    entirely    distinct    from    each    other — viz,,    the 

kow  diyid«d.    letters  that  are  used  to  designate  the  quantities  which 

are  the  subjects  of  the  science,  and   the  signs  which 

are  employed  to  indicate  certain  operations  to  be  per- 

Aigebraic       formed  on    those  quantities.      We   have   seen    that   all 

processes  : 

the  algebraic  processes  are  comprised  under  addition, 
thei:  number,   subtraction,  multiplication,  division,  and  the  extraction 
Do  not  change  of  roots  ;   and  it  is   plain,  that  the  nature  of  a  quan- 
6  uantuiee    ^^  *s  no'  ^  a^  cnange(l  by  prefixing  to  it  the  sign 
which   indicates  either  of  these  operations.     The  quan- 
tity denoted  by  the  letter  a,  for  example,  is  the  same, 
in  every  respect,  whatever  sign  may  be  prefixed  to  it ; 
that  is,  whether  it  be  added  to  another  quantity,  sub- 
tracted from  it,  whether  multiplied  or  divided   by  any 
number,  or  whether  we  extract  the  square  or  cube  or 
Algebraic       an7  other  root  of  it.     The  algebraic  signs,   therefore, 
Slgns:        must   be  regarded    merely  as   indicating   opera/tons   to 

how  regarded. 

be   performed   on    quantity,    and   not   as   affecting   the 

nature  of  the  quantities  to  which  they  may  be  prefixed. 

Plus  and       "\ye   saVj  indeed,  that  quantities   are   plus   and   minus, 

Minni. 

but  this   is   an   abbreviated  language   to  express   that 
they  are  to  be  added  or  subtracted. 

Principles  of         j£.  In   Algebra,  as   in  Arithmetic  and    Geometry 

the  science- 

From  -what  all  the  principles  of  the  science  are  deduced  from  tm 
definitions  and  axioms ;  and  the  rules  for  performing 
the  operations  are  but  directions  framed  in  conformity 

Example.  to  such  principles.  Having,  for  example,  fixed  b} 
definition,  the  power  of  thr  minus  sign,  viz.,  that  an; 


INTRODUCTION.  11 

quantity  before  which  it  is  written,  shall  be  regarded 

as  to  be  subtracted  from  another  quantity,  we  wish  to  What  *'e  wiak 

to  discover 

discover  the  process  of  performing  that  subtraction,  so 
as  to  deduce  therefrom  a  general  formula,  from  which 
we  can  frame  a  rule  applicable  to  all  similar  cases. 


SUBTRACTION. 

13.    Let  it  be   required,  for   example,  to  subtract    Subtraction. 

from  I)  the  difference  between  a  and  c. 

b 
Now,   having  written    the    letters,  with  Process. 

a  —  c 

their  proper  signs,  the  language  of  Al- 
gebra expresses  that  it  is  the  difference  only  between 
a  and  c,  which  is  to  be  taken  from  b ;  and  if  this  dif-     Difference, 
ference  were  known,  we  could  make  the  subtraction  at 
once.     But  the  nature  and  generality  of  the  algebraic 
symbols,  enable  us  to  indicate  operations,  merely,  and     Operations 

indicated. 

we  cannot  in  general  make  reductions  until  we  come 
to  the  final  result.  In  what  general  way,  therefore, 
can  we  indicate  the  true  difference  ? 


If  we   indicate   the  subtraction  of  a 
from  b,  we   have  b  —  a ;   but  then  we 


b  —  a 

b  —  a  -\-  c 


Final  formula 


have  taken  away  too  much  from  b  by 
the  number  of  units  in  c;  for  it  was  not  a,  but  the  dif- 
ference between  a  and  c  that  was  to  be  subtracted 
from  b.  Having  taken  away  too  much,  the  remainder 
is  too  small  by  c:  hence,  if  c  be  added,  the  true  re- 
mainder will  be  expressed  by  b  —  a  +  c. 

Now,  by  analyzing  this  result,  we  see  that  the  sign     Analysis  of 
of  every  term  of  the  subtrahend  has   been  changed ; 
and  what  has  been  si  own  with  respect  to  these  quan- 


12 


INTRODUCTION. 


Generalize     tities  is  equally  true  of  all  others  standing  in  the  same 

tiott. 

relation :   hence,  we   have   the   following   general   rule 
for  the  subtraction  of  algebraic  quantities  : 

Change  the  sign  of  every  term  of  the  subtrahend,  or 
Rnl«,        conceive  it  to  be  changed,  and  then  unite  the  quantities 
as  in  addition. 


Multiplica- 
tion. 


Signification 

of  the 

language, 


MULTIPLICATION. 

14.  Let  us  now  consider  the  case  of  multiplication, 
and  let  it  be  required  to  multiply  a  —  b  by  c.  The 
algebraic  language  expresses  that  the 
difference  between  a  and  b  is  to  be 
taken  as  many  times  as  there  are 
units  in  c.  If  we  knew  this  differ- 


a-b 


ac  —  be 


Process. 


ence,  we  could  at  once  perform  the  multiplication. 
But  by  what  general  process  is  it  to  be  performed 
without  finding  that  difference  ?  If  we  take  a,  c  times, 
the  product  will  be  ac ;  but  as  it  was  only  the  differ- 
ence between  a  and  6,  that  was  to  be  multiplied  by  c, 
this  product  ac  will  be  too  great  by  b  taken  c  times ; 
that  is,  the  true  product  will  be  expressed  by  ac  —  be: 
hence,  we  see,  that, 

If  a  quantity  having  a  plus  sign  be  multiplied  by 
another  quantity  having  also  a  plus  sign,  the  sign  of 
the  product  will  be  plus ;  and  if  a  quantity  having  a 
minus  sign  be  multiplied  by  a  quantity  having  a  plus 
sign,  the  sign  of  the  product  will  be  minus. 

e«jierai  case :         15.  Let  us  now  take  the  most   general  case,  viz., 
that  in  which  it  is  required  to  multipy  a  —  b  by  c  -  d. 


It»  nature. 


Principle  for 
the  signs. 


INTRODUCTION. 


13 


Let  us  again  observe  that  the  algebraic  language 
denotes  that  a  —  b  is  to  be  ta- 
ken as  many  times  as  there 
are  units  in  c  —  d ;  and  if  these 
two  differences  were  known, 
their  product  would  at  once 
form  the  product  required. 

First :  let  us  take  a  —  b  as 


a  —  b 

c-d 
etc  —  be 

—  ad  -f-  bd 
ac  —  be  —  ad  +  bd 


Iti  form. 


First  step. 


many   times    as    there   are   units   in   c ;    this   product, 

from  what  has  already  been  shown,  is  equal  to  etc  —  be. 

But  since  the  multiplier  is  not  c,  but  c  —  d,  it  follows 

that  this  product  is  too  large  by  a  —  b  taken  d  times ; 

that   is,   by  ad  —  bd :   hence,  the  first   product  dimin-    Second  step: 

ished  by  this  last,  will  give  the  true  product     But,  by 

the   rule   for   subtraction,    this   difference   is   found   by    HOW  taken. 

changing  the  signs  of  the  subtrahend,  and  then  uniting 

all  the  terms  as  in  addition :   hence,  the  true  product 

is  expressed  by  ac  —  be  —  ad  +  bd. 

By  analyzing  this  result,  and  employing  an  abbre-     Analysis  of 

the  remit. 

viated   language,  we  have  the  following  general  prin- 
ciple to  which  the  signs  conform  in  multiplication,  viz. : 


Plus  multiplied  by  plus  gives  plus  in  the  product  ; 
plus  multiplied  by  minus  gives  minus ;  minus  mul- 
tiplied by  plus  gives  minus ;  and  minus  multiplied  by 
minus  gives  plus  in  the  product. 


General 
Principle 


16.  The  remark  is  often  made  by  pupils  that  the 
above  reasoning  appears  very  satisfactory  so  long  as 
the  quantities  are  presented  under  the  above  form ; 
but  why  will  —  3  multiplied  by  —  '  give  plus  bd? 


Remark. 


Particulu 
cast. 


14  INTRODUCTION. 

How  can  the  product  of  two  negative  quantities  stand' 
mg  alone  be  plus  ? 

Minus  sign.  In  the  first  place,  the  minus  sign  being  prefixed  to 
•>  and  d,  shows  that  in  an  algebraic  sense  they  do  not 

to  interpre-  Titand  by  themselves,  but  are  connected  with  other  quan- 
*V°n'  tities ;  and  if  they  are  not  so  connected,  the  minus 
sign  makes  no  difference ;  for,  it  in  no  case  affects  the 
quantity,  but  merely  points  out  a  connection  with  other 
quantities.  Besides,  the  product  determined  above, 
being  independent  of  any  particular  value  attributed 

Foim  *f  the  *°  the  letters  a,  b,  c,  and  d,  must  be  of  such  a  form  as 
product :  ^Q  ^e  ^rue  £or  ajj  values  :  and  hence  for  the  case  in 

must  be  true 

lot  quantitiei   which  a  and  c  are  each  equal  to  zero.     Making  this 

of  any  value 

supposition,  the  product  reduces  to  the  form  of  -4-  bd. 
signs  in       The  rules  for  the  signs  in  division  are  readily  deduced 
from  the  definition  of  division,  and  the  principles  al- 
ready laid  down. 

ZERO     AND      INFINITY. 

Zero  and  17.    The  terms  zero  and  infinity  have  given  rise  to 

infinity.  much  discussion,  and  been  regarded  as  presenting  diffi- 
culties not  easily  removed.  It  may  not  be  easy  to 
frame  a  form  of  language  that  shall  convey  to  a  mind, 
id«as  not  but  little  versed  in  mathematical  science,  the  precise 
ideas  which  these  terms  are  designed  to  express  ;  but 
we  are  unwilling  to  suppose  that  the  ideas  themselves 
are  beyond  the  grasp  of  an  ordinary  intellect.  The 
terms  are  used  to  designate  the  two  limits  of  Space 
and  Number. 

18.  Assuming  any  two  points  in  space,  and  joining 


INTRODUCTION. 


15 


them  by  a  straight  line,  the  distance  between  the  points 
will  be  truly  indicated  by  the  length  of  this  line,  and 
this  length  may  be  expressed  numerically  by  the  num- 
ber of  times  which  the  line  contains  a  known  unit.  If 
now,  the  points  are  made  to  approach  each  other,  the  illustration, 

showing  th* 

length  of  the    line  will  diminish   as    the   points    come     meaning  of 

nearer  and  nearer  together,  until  at  length,  when  the  * 

two   points    become    one,   the   length   of   the   line   will 

disappear,    having    attained   its    limit,    which   is    called 

iero.     If,  on  the  contrary,  the  points  recede  from  each 

other,  the   length  of  the   line  joining   them   will   con- 

tinually  increase  ;   but   so   long   as   the   length  of  the    illustration, 

,.  ,     .  f         ,  ,.      showing  the 

line  can  be  expressed   in   terms  of  a  known   unit  of     meamngof 
measure,  it   is   not   infinite.     But,  if  we   suppose   the 


Infinity. 

points  removed,  so  that  any  known  unit  of  measure 
would  occupy  no  appreciable  portion  of  the  line,  then 
the  length  of  the  line  is  said  to  be  Infinite. 

19.  Assuming  one  as  the  unit  of  number,  and  ad- 
mitting the  self-evident  truth  that  it  may  be  increased 
or  diminished,  we  shall  have  no  difficulty  in  under- 
standing the  import  of  the  terms  zero  and  infinity,  The  term* 

.       Zero  and  In- 

as  applied   to  number.     For,  if  we  suppose   the   unit    fin;ty  applied 

one  to  be  continually  diminished,  by  division  or  other-    1 

wise,  the  fractional  units  thus  arising  will  be  less  and    niu*tratio». 

less,  and   in   proportion    as  we  continue  the  divisions, 

thfy    will   continue   to   diminish.      Now,   the    limit   or 

boundary  to  which  these  very  small  fractions  approach, 

is  called   Zero,  or  nothing.     So  long  as  the  fractional         Zeio: 

number   forms    an    appreciable  part  of  one,  it   is   not 

zero,  but  a  finite  fraction  ;  and  the  term  zero  is  only 


16  INTRODUCTION. 

applicable  to  that  which  forms  no  appreciable  part  of 
the  standard. 

illustration.  If,  on  the  other  hand,  we  suppose  a  number  to  be 
continually  increased,  the  relation  of  this  number  to  the 
unit  will  be  constantly  changing.  So  long  as  the  num- 
ber can  be  expressed  in  terms  of  the  unit  one,  it  is 
infinity ;  finite,  and  not  infinite  ;  but  when  the  unit  one  forms 
no  appreciable  part  of  the  number,  the  term  infinite 
is  used  to  express  that  state  of  value,  or  rather,  that 
limit  of  value. 

The  terms,         20.   The  terms   zero  and  infinity  are  therefore  em- 
ployed to  designate  the  limits   to  which  decreasing  and 

employed. 

increasing  quantities  may  be  made  to  approach  nearer 
A»  limiu.     than  any  assignable  quantity ;   but  these  limits  cannot 
be  compared,  in  respect  to  magnitude,  with  any  known 
standard,  so  as  to  give  a  finite  ratio. 

Why  limits?  21.  It  may,  perhaps,  appear  somewhat  paradoxical, 
that  zero  and  infinity  should  be  defined  as  "  the  limits 
of  number  and  space"  when  they  are  in  themselves 
not  measurable.  But  a  limit  is  that  "  which  sets  bounds 

Definition  of  to,  or  circumscribes ;"  and  as  all  finite  space  and  finite 
number  (and  such  only  are  implied  by  the  terms  Space 
pace  a  an(j  Dumber),  are  contained  between  zero  and  infinity, 
we  employ  these  terms  to  designate  the  limits  of  Num- 
ber and  Space. 

OF     THE     EQUATION. 

Subject  of          22.    The  subject  of  equations   is   divided   into   two 
.quatious :  tg      The  fa.   consists  in  finding  the  equation  ;  that 

ho\r  divided.     r 

First  part:     is,  in  the  process  of  expressing  the  relations  existing 


INTRODUCTION. 


17 


between  the  quantities  considered,  by  means  of  the 
algebraic  symbols  and  formula.  This  is  called  the 
Statement  of  the  proposition.  The  second  is  purely  statement. 

Second  part. 

deductive,   and   consists,  in  Algebra,   in  what  is   called 
the   solution  of  the   equation,  or  finding  the  value   of       Solution, 
the  unknown  quantity ;  and  in  the  other  branches  of 
anah-sis,  it  consists   in  the  discussion  of  the  equation ;    ] 

'  an  equati«« 

that  is,  in  the  drawing  out  from  the  equation  every 
proposition  which  it  is  capable  of  expressing. 

23.   Making  the   statement,  or  finding  the  equation,     Statement: 
is    merely  analyzing   the    problem,  and  expressing   its      what  it  u. 
elements  and  their  relations  in  the  language  of  analy- 
sis.     It    is,   in    truth,  collating   the    facts,   noting   their 
bearing   and    connection,   and    inferring    some   general 
law  or  principle   which    leads   to  the   formation  of  an 
equation. 

The    condition    of  equality   between    two    quantities     Equality  of 

two  quanti- 

is  expressed  by  the  sign  of  equality,  which  is  placed 
between    them.     The  quantity  on   the  left  of  the  sign 
of  equality  is   called   the   first   member,   and   that   on    lst  member, 
the  right,  the  second  member  of  the  equation.     Hence,    2d  member- 
an  equation   is   merely  a   proposition   expressed   aJge-     Proposition, 
braically,  in  which  equality  is  predicated  of  one  quan- 
tity as  compared  with  another.     It  is  the  great  formula 
of  Algebra. 


ties: 
Hour  ex 

pressed. 


24.  Every  quantity  is  either  abstract  or  concrete  : 
hence,  an  equation,  which  is  a  general  formula  for 
expressing  equality,  must  be  either  abstract  or  con- 
crete. 


Abstract. 
Conci  etc. 


18 


INTRODUCTION. 


Abstract 

»q  nation. 


Concrete 
equation. 


Five  opera- 
tions may  be 
y#rformed. 


A  ti  em*. 
First. 


Second. 


ThM. 


An  abstract  equation  expresses  merely  the  relation 
of  equality  between  two  abstract  quantities :  thus, 

a  -\-  b  =:  x, 

is  an  abstract  equation,  if  no  unit  of  value  be  assigned 
to  either  member ;  for,  until  that  be  done  the  abstract 
unit  one  is  understood,  and  the  formula  merely  ex- 
presses that  the  sum  of  a  and  b  is  equal  to  x,  and  is 
true,  equally,  of  all  quantities. 

But  if  we  assign  a  concrete  unit  of  value,  that  is, 
say  that  a  and  b  shall  each  denote  so  many  pounds 
weight,  or  so  many  feet  or  yards  of  length,  x  will  be 
of  the  same  denomination,  and  the  equation  will  be- 
come concrete  or  denominate. 

25.  We  have  seen  that  there  are  five  operations 
which  may  be  performed  on  an  algebraic  quantity 
(Art.  3).  We  assume,  as  an  axiom,  that  if  the  same 
operation,  under  either  of  these  processes,  be  performed 
on  both  members  of  an  equation,  the  equality  of  the. 
members  will  not  be  changed.  Hence,  we  have  the 
five  following 

AXIOMS. 

1.  If  equal   quantities   be   added   to  both   members 
of  an  equation,  the  equality  of  the  members   will  not 
be  destroyed. 

2.  If  equal  quantities  be  subtracted  from  both  mem- 
bers of  an  equation,  the  equality  will  not  be  destroyed. 

3.  If  both  members  of  an  equation  be  multiplied  by 
the  same   number,  the  equality  will  not  be  destroj  ed 


INTRODUCTION.  19 

4.  If  both  members  of  an  equation  be  divided  by  Fourth, 
the  same  number,  the  equality  will  not  be  destroyed. 

5.  If  both  members  of  an  equation  be  raised  to  Fifth, 
the  same  power,  the  equality  of  the  members  will 

not  be  destroyed. 

6.  If  the  same  root  of  both  members  of  an  equa-  sixth, 
tion  be  extracted,  the  equality  of  the  members  will 

not  be  destroyed. 

Every  operation   performed  on   an  equation  will      Use  of 
fall  under  one  or  other  of  these  axioms,  and  they 
afford  the  means  of  solving  all  equations  which  ad- 
mit of  solution. 

26t   Two   quantities  are  said  to  be  equal,  when  Equality  de- 
fined, 
each  contains  the  same  unit   an   equal  number  of 

times.    Hence,  the  term  equal  applies  to  measures, 

and  has  the   same   signification   in  Arithmetic,  in  E(iual  to  *& 

parts. 

Algebra,  and  in  Geometry.  If,  in  Geometry,  two 
figures  can  be  applied  to  each  other,  so  as  to  coin- 
cide or  fill  the  same  space,  they  are  said  to  be 
equal  in  all  their  parts. 

27t  We  have  thus  pointed  out  some  of  the  marked 
characteristics  of  Algebra.      In  Algebra,  the  quan-    Classes  of 

quantities  in 

tities,  about  which  the  science  is  conversant,  are  Algebra, 
divided,  as  has  been  already  remarked,  into  known 
and  unknown,  and  the  connections  between  them, 
expressed  by  the  equation,  afford  the  means  of 
tracing  out  further  relations,  and  of  finding  the 
values  of  the  unknown  quantities  in  terms  of  the 
known. 


20  INTRODUCTION. 

SUGGESTIONS     FOR     THOSE     WHO     TEACH    ALGEBRA. 

utters  are  but       j_  ^  careful  to  expiain  that  ^  ietters  employed. 

mere  gymtnls  *•      *, 

are  the  mere  symbols  of  quantity.  That  of  and  in  them- 
selves, they  have  no  meaning  or  signification  whatever, 
but  are  used  merely  as  the  signs  or  representatives 
of  such  quantities  as  they  may  be  employed  to  denote. 
Signs  indicate  2.  Be  careful  to  explain  that  the  signs  which  are 

operations. 

used  are  employed  merely  for  the  purpose  of  indicating 
the  five  operations  which  may  be  performed  on  quan- 
tity ;  and  that  they  indicate  operations  merely,  without 
at  all  affecting  the  nature  of  the  quantities  before  which 
they  are  placed. 

Letter*  and         3     Explain    that   the  letters    and  signs   are  the  ele- 
uiji'*  elements 

of  language,    ments  of  the  algebraic  language,  and  that  the  language 

itself  arises  from  the  combination  of  these  elements. 
Algebraic  4    Explain  that  the  finding  of  an  algebraic  formula 

formula 

is  but  the  translation   of  certain   ideas,  first  expressed 

in  our  common  language,  into  the  language  of  Algebra; 

its  interpreta-   anf]  that  the  interpretation  of  an  algebraic  formula  is 

tion. 

merely  translating  its  various  significations  into  common 

language. 
Language.          5.  Let  the  language  of  Algebra  be  carefully  studied, 

so  that  its  construction  and  significations  may  be  clearly 

apprehended. 

Co-efficient.         6.   Let  the   difference  between  a  co-efficient  and  an 
;  x  oncnt      exponent  be  carefully  noted,  and  the  office  of  each  often 

explained ;  and  illustrate  frequently  the  signification  of 

the  language  by  attributing  numerical  values  to  letlers 

in  various   algebraic  expressions. 

7.  Point  out  often  the  characteristics  of  similar  and 


INTRODUCTION.  21 

dissimilar  quantities,  and  explain  which  may  be  incor-       similar'' 

quantities' 

porated  and  which  cannot. 

8.  Explain  the  power  of  the  minus  sign,  as  shown    Minus  sie« 
in  the  four  ground  rules,  but  very  particularly  as  it  is 
illustrated  in  subtraction  and  multiplication. 

9.  Point  out   and    illustrate    the  correspondence  be-     Arithmetic 

and  Algebra 

tween  the  four  ground   rules  of  Arithmetic  and  Alge-      compared. 
bra  ;  and  impress  the  fact,  that  their  differences,  where- 
ever    they   appear,   arise    merely   from  *  differences   in 
notation   and   language :    the    principles    which    govern 
the  operations  being  the  same  in  both. 

10.  Explain  with  great  minuteness  and  particularity,      Equation. 
all    the   characteristic   properties  of  the   equation ;    the    Its  properties- 
manner  of  forming  it ;  the  different  kinds  of  quantity 

•which    enter   into  its  composition  ;    its  examination  or 
discussion  ;  and  the  different  methods  of  elimination. 

11.  In  the  equation  of  the  second  degree,  be  careful    Equation  oi 

the  second 

to  dwell  on  the  four  forms  which  embrace  all  the  cases,        degree. 

and   illustrate  by  many  examples    that  every  equation 

of  the  second  degree  may  be  reduced  to  one  or  other 

of  them.      Explain   very  particularly  the  meaning  of       its  form. 

the  term  root ;  and  then  show,  why  every  equation  of       ite  root». 

the  first   degree   has   one,  and   every   equation   of  the 

second  degree  two.     Dwell  on  the  properties  of  these 

roots  in  the  equation  of  the  second  degree.     Show  why 

their  sum,  in  all  the  forms,  is  equal   to  the  co-efficient     Th«ir*nm. 

of  the  second   term,  taken  with   a  contrary  sign  ;  and 

why  their  product  is  equal  to  the  absolute  term  with  a  Their 

contrary  sign.      Explain  when  and  why  the  roots  are 

imaginary. 


22  INTRODUCTION. 

General  12.  In  fine,  remember  that  every  operation  and  rule 

is  based  on  a  principle  of  science,  and  that  an  intelli- 
gible reason   may  be  given  for  it.     Find  that  reason, 
and  impress  it  on  the  mind  of  your  pupil  in  plain  and 
should  be      simple  language,  and  by  familiar  and  appropriate  illus- 
trations.    You  will  thus  impress  right  habits  of  inves- 
tigation   and   study,  and   he   will   grow  in    knowledge. 
The   broad    field   of    analytical    investigation    will    be 
opened    to   his   intellectual    vision,   and    he    will    have 
made  the  first  steps  in  that  sublime  science  which  dis- 
They  lead  to    covers  the  laws  of  nature  in  their  most  secret  hiding-' 
if«nerai lawSl    places,  and  follows  them,  as  they  reach  out,  in  omnipo- 
tent power,  to  control  the  motions  of  matter  through 
the   entire   regions   of  occupied    space. 

(See    Davies'  Nature  and    Utility   of    Mathematics, 
Article  Algsbra). 


KEY 


EQUATIONS  OF  THE  FIRST  DEGREE. 

~.  5x      4x      10      7      13z 

1.  Given  ____  13  =  --.—. 

VERIFICATION. 

5  x  11.1      4  x  11.1  7      13  x  11.1 

12  3  "86 

Multiply  by  24,  least  common  multiple, 
10x11.1  —  32  x  11.1  —  312  =  21  —  52  x  11.1;  that  is, 
—  556.2=  —556.2. 

2.  Given  x  -f  18  =  3z  —  5,  to  find  x. 
Transposing  and  reducing, 

-  2ar  =  -  23  ; 
dividing  both  members  by  —  2, 

x=  11J. 

3.  Given        6  -  2x  +  10  =  20  —  3*  —  2,  to  find  *. 
Transposing  and  reducing, 

x  =  2. 

4.  Given  ar"^"Hz"^o  *  =  H>to  nn^  #• 

2         o 

Multiplying  both  members  by  6,  and  reducing, 

11*  =  66; 
whence,  x  =  6. 


24  KEY  TO  DAVIES'  BOURDON.  [91. 

5.  Given  2z  —  -  x  +  1  =  5x  —  2,  to  find  x. 

tt 

Multiplying  beth  members  by  2,  transposing  and  reducing, 
—  Tx  =  -6; 

whence,  x  =  -. 

6.  Given          Box  +  -    -  3  =  bx  —  a,  to  find  a. 

£ 

Multiplying  by  2,  transposing  and  reducing, 

Qax  —  2bx  =  6  —  3a ; 

factoring  the  first  member  of  the  equation,  we  have 
(6a  —  26)  x  —  6  —  3a ; 
6  —  3a 


whence,  a;  = 


6a  -  26' 


7.  Given  ?— i  +  |  =  20  -  ^^,  to  find  *, 

SI          8  /* 

Multiplying  both  members  by  6, 

Bx  -  9  +  2*  =  120  -  3  x  +  57 ; 
transposing  and  reducing, 

8*  =  186  ;  .  • .  «  =  23£. 

a:  +  3       a;              a;  —  5 
3.  Given         — h  «  =  4 — ,  to  find  x. 

Multiplying  both  members  by  12. 

Gx  +  18  +  4x  =  48  -  3ar  4-  15 ; 
transposing  and  reducing, 

13*  =  45  ..  • .  t 


EQUATIONS  OF  THE  FIRST  DEGREE.  25 

ax  —  b       a       bx       bar  —  a 

9.  Given       —  —  --  h  g  =  -^  --  g—  ,  to  find  x. 

Multiplying  both  members  by  12, 

Sax  —  Sb+4a  =  Gbx  —  4bx  +  4a; 
transposing,  reducing  and  factoring, 

36 
(3a  —  26)  x  =  36,  .  •  .  x  — 

Sax      %bx 

10.  Given  ---  ;  --  4  =/,  to  find  x. 

c          a 

Multiplying  both  members  by  cd, 

3adx  —  2bcx  —  4cd  =fcd  ; 
transposing,  reducing  and  factoring, 


3a  — 


8aa;  —  b       36  —  c 

11.  Given  —  -  ---  s  —  =  4  —  6,  to  find  x. 

i  '-~ 

Multiplying  both  members  by  14, 

IQax  -  26  -  216  +  7c  =  56  -  146  ; 

transposing  and  reducing, 

56  +  «ft 


16a 


12.  Given  -  —  — 1-  -  =  —,  to  find 

Multiplying  both  members  by  30, 

Qx  -  Wx  +  20  +  15  x  =  130 ; 
transposing  and  reducing, 

lla;  =  110;  „•.  x  =  10. 


KEY  TO  DAVIES'  BOURDON.  [91-92. 


.„,-,.  X        X    .    X        X 

13.  Given 


T  ^ -j  =/,  to  find  x. 

abed 

Multiplying  both  members  by  abed,  and  factoring, 
(bed  —  acd  +  abd  —  abc)  x  =abcdf  .  • .  x  = 


bed  — acd  -j-  abd  —  abc 


ij    n-  8* —5  .   4* -2  ,    ,         .    , 

14.  Given     a; 1 — —  =  x  +  1,  to  find  x. 

lo  1 1    . 

Multiplying  both  members  by  143, 

143*  -  33*  +  55  +  52x  —  26  =  143*  +  143 ; 
transposing  and  reducing, 

19*  =  114;  •.  *s_-6. 

15.  Given    y  -  y  -  ^^  =  -  12f|,  to  find  *. 

Multiplying  both  members  by  315, 

45x  —  280*  -  63*  -|-  189  =  —  3983  ; 
transposing  and  reducing, 

-298*  =-4172;  .-.  *  =  14. 

4*  -  2      3*  -  1   t 

16.  Given       2* —  =  — - — ,  to  find  *. 

O  ii 

Multiplying  both  members  by  10, 

20*  —8*+  4  =  15ar  — 5; 
transposing  and  reducing, 

—  3*  =  -  9 ;  .  • .  x  =  3. 


92.]  EQUATIONS  OF  THE  FIRST  DEGREE.  27 

17.  Given       Sx  -\ —  =  x  +•  a,  to  find  a?. 

o 

Multiplying  both  members  by  3, 

9x  +  bx  —  d  =  Sx  -f  3a  ; 
transposing,  reducing  and  factoring, 

O_      I       J 

(6  -f  b)  x  =  3a  -f  d  ;  .  • .  x  = 


18.  Given 


a  —  o  a  -4-  b  b 

to  find  x. 

We  see  that  the  least  common  multiple  of  the  several 
fractions  of  the  two  members  of  this  equation  is, 

a*b  —  b5. 

Hence,  multiplying  both  members  of  the  equation  by 
a?b  —  b3,  and  performing  all  the  indicated  operations,  we 
shall  have, 

a?bx  +  2ab*x  +  t>*z  —  aW  —  2ab*  —  54  —  3asb  +  3«i3  =  4a2^ 
-  5ab3  +  J4  —  Zatbx  +  Wx  +  a4  —  a?bx  —  tf&  +  IPx  ; 


then,  by  transposing, 

bsx  +  2a2Ja;  —  aj3^  +  a?bx  —  b*x  =  4a252  — 

J4 


factoring,  we  have, 

2J  (2a2  +  a5  —  J2)  a;  =  a4  +  3a3b 
dividing  by  the  coefficient  of  x, 
3asb 


2b  (2«2  +  ab  - 


'28  KEY    TO    DAVIES5    BOUKDON.  [92. 

19.  Given,  x  =  3x  —  \ (4  —  x)  +  ^ . 

iii  O 

Clearing  of  fractions,  and  dropping  parenthesis, 

6x  =  ISx  —  12  +  3x  +  2. 
Transposing  and  reducing, 

-I5x=  —  10; 
2 

•  o»     .....  

3x  —  7      25  —  4z      5a;  —  14 

20.  Given,        -fr-  +  —^-~  =  -^— . 

Clearing  of  fractions, 

27a;  —  63  +  125  —  20z  =  75a;  —  210. 
Transposing  and  reducing, 

_  682;  =—272; 
x  =  4. 

2z  +  5      40  —  x      Wx  —  427 

21.  Given,     -^  +  -_          -J^-. 

Clearing  of  fractions, 

304a;  +  760  +  9880  —  247z  =  1040z  —  44408. 
Transposing  and  reducing, 

—  983z  =  —  55048 ; 
x  =  56. 

00~.  x      x  —  5  /2«11\ 

22.  Given, —  +  5  =  2;-^—  +  ij . 

Clearing  of  fractions, 

Ux  —  7x  +  35  +  385  =  77z  —  2»  —  77. 
Transposing  and  reducing, 

—  71z=  —497; 
a;  =7. 

a;  —  1       »  — 2      a;  +  3      z  +  4 

23.  Given,     -^-  +  -^-=-±-  +  -^  +  1. 

Clearing  of  fractions, 

6x  —  6  +  4z  —  8  =  3x  +  9  +  2.r  +  8  +  12. 


92.]  EQUATIONS    OF   THE   FIRST   DEGREE.  29 

Transposing  and  reducing, 

5z  =  43 ; 
«  =  8f 

X  ~  l       X  ~  2       X  ~  5       a?~6 


24.  Given, 

Performing  the  indicated  subtraction  in  both  members, 

—  1  —1 

(x  _  2)  (z  —  3)  ~~  (x  —  6)  (z  —  7) ' 

Clearing  of  fractions,  and  performing  indicated  operations, 

Transposing  and  reducing, 

—  8x  =  —  36 ; 
#  =  4i. 


25.  Given,      *  +        aj  -       -  (*  +  5)  (x-  3)  +     =  0. 
Performing  indicated  operations, 

s?  +  x-.15-tf-2x  +  15  +  ^  =  0, 

Transposing  and  reducing, 

—  x  =—12;    .-.    3  =  12. 

6z  +  7      2rc  —  2      2x  +  1 


26.  Gwen, 


Clearing  of  fractions, 

42^  +  I3x  —  42  —  30z  +  30  =  42a^  —  15a;  —  18. 
Transposing  and  reducing, 

—  2x  =  —  6  ; 
x  =  3. 

/*  i  •  Lriven,  —  —          —  ^^         —  —         ~  • 

x  —  2      x  —  4      x  —  6      x  —  8 

Performing  the  indicated  subtractions, 

—  2  _  —2 

(x  —  2)  (x  —  4)  ~  (x  —  6)  (x  —  8)  ' 


3  KEY  TO   DAVIES'   BOURDON.  [92. 

Dividing  by  —  2,  and  clearing  of  fractions, 
x2  —  14z  +  48  =  tf  —  6x  +  8. 
Transposing  and  reducing, 

—  Sx=  —  40; 
x  =  5. 

28.  Given,          (x  +  I)2  =  (5  +  x)  x  —  2. 
Performing  indicated  operations, 

z3  +  2x  +  1  =  5x  +  a?  —  2. 
Transposing  and  reducing, 

—  3z  =  —  3; 

3  =  1. 

29.  Given,          5-^-=-  +  -^-  =  g-^—  . 

2x  —  5      x  —  3      3^  —  1 

Clearing  of  fractions, 

6«2  —  20a;  +  6  +  6z2  —  llx  +  5  =  12o«  —  66a  +  90. 
Transposing  and  reducing, 


_79 
~' 


30.  Given,  -  +     *  a 


a      b  —  a      b  -\-  a* 

Factoring, 

jl   ,       1      >          a 

/»    ^  I      ,  \    __  -__j__ 

(o      d  —  a)~J  +  a* 
Reducing, 

b        \_        g 

aCJ  -«)/(&  +  «)' 


_ 

~ 


•fa)' 

. /       n\      i  / 
81.  Given,    -| 


93-97.]  EQUATIONS   OF   THE   FIRST    DEGEEE.  31 

Performing  indicated  operations, 


1          1 
X  ~a 


Clearing  of  fractions,  and  reducing, 

=  8a  ; 


32.  Given,      1.2*  -  'lSx  ~  >05  =  .4*  +  8.9. 

.5 

Clearing  of  fractions, 

.6*  —  .18*  +  .05  =  3n  +  4.45. 
Transposing,  and  reducing, 


a;  =  20. 

_   AK 

33.  Given,      4.8z  -  '  =  1.6*  +  8.9. 

Clearing  of  fractions, 

2.4z  —  .72*  +  .05  =  0.8z  +  4.45. 
Transposing  and  reducing, 

.882;  =  4.40; 
*  =  5. 


STATEMENT  AND  SOLUTION  OP  PROBLEMS. 

8.  Divide  $1000  between  A,  B,  and  C,  so  that  A  shall  have 
£72  more  than  B,  and  C  $100  more  than  A. 

Let       x        denote  the  number  of  dollars  in  B's  share. 
Then  will  *  +  72  "  «  «  "      A's     " 

and  *  +  72  +  100 "  "  «  "      C's     " 

From  the  conditions  of  the  problem, 


32  KEY  TO  DAVIES'  BOUKDON.  [37. 

x  +  x  +  72  +  x  +  172  =  1000;     or,     3  x  =  756  ,     .  •  .    x  -=  252, 
or,         A's  share  is  $324,  B's  share  $252  and  C's  share  $424. 

9.  A  and  B  play  together  at  cards.     A  sits  down  with  $84  and  B 
with  $48.     Each  loses  and  wins  in  turn,  when  it  appears  that  A  has 
five  times  as  much  as  B.     How  much  did  A  win? 

Let  x  denote  the  number  of  dollars  that  A  wins. 
Then  will  84  +  x  denote  what  A  has  at  last, 
and  48  —  x  what  B  has  at  last  ; 

from  the  conditions  of  the  problem, 

84  +  *  =  5  (48  —  x)  ;         or,         84  +  x  =  240  -  5x  ; 
whence,  ?•  =  26  ;         or,         A  wins  $26. 

10.  A  person  dying,  leaves  half  of  his  property  to  his  wife,  one 
sixth  to  each  of  two  daughters,  one  twelfth  to  a  servant,  and  the 
remaining  $GOO  to  the  poor  :   what  was  the  amount  of  his  property  ? 

Let          x  denote  the  whole  number  of  dollars  in  the  property. 
Then  will  ^       "  "  "  "         "  in  the  wife's  share. 

it 

-       "  "  «  "         «  each  daughter's  '' 

6 

and  -^      "  "  "  "        "  the  servant'?       " 

1  £ 

from  the  conditions  of  the  problem, 


multiplying  both  members  by  12,  transposing  and  reducing, 
-  x  =  -  7200  ;         or,         x  =  7200. 


97.]  EQUATIONS  OF  THE  FIKST  DEGREE.  33 

1 1 .  A  father  leaves  his  property,  amounting  to  $2520,   to  four 
sons,  A,  B.  C  and  D.     C  is  to  have  $360,  B  as  much  as  C  and  D 
together,  and  A  twice  as  much  as  B  less  $1000:  how  much  do  A, 
B  and  D  receive  ? 

Let        x         denote  the  number  of  dollars  that  D  receives  • 
Then  will  x  +  360       "  "  "  "   B        " 

and         2* +  720 -1000       "  "  "   A        " 

from  the  conditions  of  the  problem, 

360  -(-  x  -f  x  +  360  -f  2x  +  720  -  1000  =  2520; 
transposing  and  reducing, 

4x  =  2080  ;  .  • .  x  =  520 

or,     D's  share  is  $520  ;  B's  share  $880,  and  A's  share  $760. 

12.  An  estate  of  $7500  is  to  be  divided  between  a  widow,  two 
sons,  and  three  daughters,  so  that  each  son  shall  receive  twice  as 
much  as  each  daughter,  and  the  widow  herself  $500  more  than  all 
the  children :  what  was  her  share,  and  what  the  share  of  each  child  ? 

Let          x  denote  the  number  of  dollars  in  each  daughter's  share  ; 
Then  will  2x        "  "  "  "         son's  " 

and  4x  +  3z  +  500"  "  "         the  widow's " 

from  the  conditions  of  the  problem, 

4x  +  3x  +  4*  +  3z  -|-  500  =  7500; 
transposing  and  reducing, 

Uas  =  7000 ;  • .  x  =  500. 

Daughters'  share  $500;  son's  share  $1000 ;  widow's  share  $4000. 

13.  A  company  of  180  persons  consists  of  men,  women  and 


34:  KEY  TO  DAVIEs'  BOURDON.  [97-98. 

children.  The  men  are  8  more  in  number  than  the  women,  and  the 
children  20  more  than  the  men  and  women  together  :  how  many  of 
each  sort  in  the  company  1 

Let  x  denote  the  number  of  women  ; 
Then  will  x  +  8    "  "  men  ; 

and  x  +  x  +  8  +  20  "  children. 

From  the  conditions  of  the  problem, 

x  +  x  +  8  +  x  +  x  +  8  +  ZO  =  lSQ; 
transposing  and  reducing, 

4x=  144;  .  •.  x  =  36. 

36  women,  44  men  and  100  children. 

14.  A  father  divides   $2000  among  five  sons,  so  that  each  elder 
should  receive  $40  more  than  his  next  younger  brother :  what  is 
the  share  of  the  youngest  ? 

Let  x  denote  the  number  of  dollars  in  the  youngest's  share. 
Then  will  x  4-  40 "  "  "          "       second's         " 

x  +  80 "  "  "          "       third's  " 

x  -f  120  "  "          "        fourth's          " 

x  +  160  "  "          "       fifth's  " 

From  the  conditions  of  the  problem, 

5*  +  400  =  2000 ; 
transposing  and  reducing, 

5z=1600;  .-.  2  =  320. 

15.  A  purse  of  $'2850  is  to  be  divided  among  three  persons,  A, 
P>  and  C ;  A's  share  is  to  be  T6T  of  B's  share,  and  C  is  to  havf  $300 
more  than  A  and  B  together  :  what   s  each  one's  share  ? 


98.]  EQUATIONS  OF  THE  FIRST  DEGEEE. 

Let         x  denote  the  number  of  dollars  in  B's  share. 

Then  will   ^     "  "  "  "      A's     " 

RX 
and  x  +  -j  +  300  «  "      C's     " 

From  the  conditions  of  the  problem, 

*  +  fl  +  *  +  if  +300  =  2850; 
clearing  of  fractions,  transposing  and  reducing, 

34z  =  28050  ;  .  • .  x  =  825  ; 

hence,  B's  $825  ;  A's  $450  ;  and  C's  $1575. 

16.  Two  pedestrians  start  from  the  same  point;  the  first  steps 
twice  as  far  as  the  second,  but  the  second  makes  five  steps  while  tho 
first  makes  but  one.  At  the  end  of  a  certain  time  they  are  300  feet 
apart.  Now.  allowing  each  of  the  longer  paces  to  be  3  feet,  how  for 
will  each  have  travelled  1 

Let          x       denote  the  number  of  feet  travelled  by  the  first. 

Then  will    •  "  "  steps  "        " 

8 

5x 

"  "  "       second. 

3 

and  11  x  §,  or  ^  "  feet          "          «        " 

O  O 

I 

From  the  conditions  of  the  problem, 

1^  -  x  =  300  ; 
o 

clearing  of  fractions,  transposing  and  reducing, 

15a: 

9z  =  1800 ;  .  • .  y  =  200     ard         r  =  500. 

0 


36  KEY  TO  DAVIES'  BOURDON. 

17.  Two  carpenters,  24  journeymen,  and  8  apprentices,  received 
at  the  end  of  a  certain  time  $144.     The  carpenters  received  $1  per 
day.  each  journeyman  half  a  dollar,  and  each  apprentice  25  cents : 
Slow  many  days  were  they  employed  1 

Let        x  denote  the  number  of  days. 

Then  will  x       "  dollars  due  each  carpenter.    - 

x 

journeyman. 
2 

x 
and  -  apprentices ; 

from  the  conditions  of  the  problem, 

reducing, 

16*  =  144 ;  .  • .  r  =  9. 

18.  A  capitalist  receives  a  yearly  income  of  $2940  ;  four  fiths  of 
his  money  bears  an  interest  of  4  per  cent.,  and  the  remainder  of  5 
per  cent. :  how  much  has  he  at  interest  ? 

Let      x     denote  the  number  of  dollars  at  interest. 

4«        4 

Then  will  —  X  -r-r^  denote  the  interest  of  1st  parcel. 
O         100 


From  the  conditions  of  the  problem, 

4x        4        x        5 

"5"XTOO+5X  T00  = 

clearing  of  fractions,  and  reducing, 

21*  =  1470000 ;  .  • .  x  =  70000, 


98.]  EQUATIONS  OF  THE  FIRST  DEGREE.  37 

19.  A  cistern  containing  60  gallons  of  water  has  three  unequal 
cocks  for  discharging  it  ;  the  largest  will  empty  it  in  one  hour,  the 
second  in  two  hours,  and  the  third  in  three  :  in  what  time  will  the 
cistern  be  emptied  if  they  all  run  together  ? 

Let  x  denote  the  required  number  of  minutes. 

Then  since  the  first  emits  1   gallon  per  minute,  the  second  \  of  a 
gallon  per  minute,  and  the  third  £  of  a  gallon, 

x  will  denote  the  number  of  gallons  emitted  by  the  1st. 

ti  U  ((  (t  (C  O(] 

2 

U  ((  («  U  it  0,1 

3 

From  the  conditions  of  the  problem, 


clearing  of  fractions  and  reducing, 

11s  =  360  .-.  x  =  321§rw. 

20.  In  a  certain  orchard  •£  are  apple-trees,  -J  peach-trees,  \  plum- 
trees,  120  cherry-trees,  and  80  pear-trees  :  how  many  trees  in  the 
orchard  ? 

Let  x  denote  the  whole  number  of  trees. 

(K 

Then  will     -      "  "  apple-trees. 

"  "  "        "  peaeh-trees. 

I      "  "  "         "plum-trees. 

From  the  conditions  of  the  problem, 


KEY  TO  DAVIES'  BOURDON.  [99. 


clearing  of  fractions,  transposing  and  reducing, 

—  x  =  -  2400  .  •  .  x  =  2400. 

21.  A  farmer  being  asked  how  many  sheep  he  had,  answered  that 
he  had  them  in  five  fields;  in  the  1st  he  had  i,  in  the  2d  £,  in  the. 
3d  £,  in  the  4th  fa,  and  in  the  5th  450  :  how  many  had  he  ? 

Let          x  denote  the  whole  number  of  sheep  : 

Then  will   -A         "  "  "  "      in  1st  field. 

4 

x 

«  u  u  <c  ^d    " 

6 


and  "  "  "  "  4th    " 

12 

From  the  conditions  of  the  problem, 


multiplying  both  members  by  24,  transposing  and  reducing, 
-  Qx  =  —  10800  .  •  .  x=  1200. 

22.  My  horse  and  saddle  together  are  worth  $132,  and  the  horse 
is  worth  ten  times  as  much  as  the  saddle  :  what  is  the  value  of  the 
horse  ' 

Let        x  denote  the  number  of  dollars  that  the  saddle  is  worth. 
Then  will  10*     "  "  "  "      horse         " 

From  the  conditions  of  the  problem, 


99.]  EQUATIONS  OF  THE  FIRST  DEGREE.  39 

x  +  10*  =  132 ; 
reducing,     llx  =  132         .'.      x  =12;     whence,     10*  =  120. 

23.  The  rent  of  an  estate  is  this  year  8  per  cent,  greater  than  it 
was  last.     This  }  ear  it  is  $1890 :  what  was  it  last  year  ? 

Let        x  denote  the  number  of  dollars  in  last  year's  rent. 
Then  will  x  +  ^  "  "  "  this     "        « 

From  the  conditions  of  the  problem, 


clearing  of  fractions  and  reducing, 

108*  =  189000  ;  .  • .  x  =  1750. 

24.  What  number  is  that  from  which,  if  5  be  subtracted,  £  of  the 
remainder  will  be  40  ? 

Let  x  denote  the  number  required : 
From  the  conditions  of  the  problem, 

t(s-5)  =  40; 

clearing  of  fractions,  performing  operations  indicated,  transposing 
and  reducing, 

2x=  130;  .  •.  x  =  65. 

25.  A  post  is  $  in  the  mud,  £  in  the  water,  and  ten  feet  above  the 
water :  what  is  the  whole  length  of  the  post  ? 

Let          x  denote  the  number  of  feet  in  length. 

Then  will    ^         '«  "  «  «          the  mud ; 

4 

and  «  «  «  «          the  water: 

o 


40  KEY  TO  DAVIES5  BOUBDON.  [99. 

From  the  conditions  of  the  problem ; 

y  sy 

4+3+10  =  *; 

clearing  of  fractions,  transposing  and  reducing, 

-  5z  =  -  120  j  .  • .  x  -  24. 

26.  After  paying  -J-  and  £  of  my  money,  1  had  66  guineas  left  in 
my  purse  :  how  many  guineas  were  in  it  at  first  1 

Let  x  denote  the  number  at  first ; 
from  the  conditions  of  the  problem, 

.-|-  =  =  Mj 

clearing  of  fractions,  transposing  and  reducing, 

llz  =  1320;  .'.  x  =  120. 

27.  A  person   was  desirous  of  giving  3  pence  apiece  to  some 
beggars,  but  found  he  had  not  money  enough  in  his  pocket  by  8 
pence ;  he  therefore  gave  them  each  two  pence  and  had  3  pence 
remaining  :  required  the  number  of  beggars. 

Let  x  denote  the  number  of  beggars ; 
then,  by  the  first  condition, 

3z  —  8         denotes  the  number  of  pence ; 
by  the  second  condition, 

2a;  -f  3         denotes  the  number  of  pence  ; 
hence,  3z  —  8  =  2ar  +  3  ; 

transposing  and  reducing 

a;  =11. 


39.]  EQUATIONS  OF  THE  FIRST  DEGREE.  41 

28.  A  person  in  play  lost  %  of  his  money,  and  then  won  3 
shillings  ;  after  which  he  lost  %  of  what  he  then  had  ;  and  this  done, 
found  that  he  had  but  12  shillings  remaining:  what  had  he  at  first1? 

•  Let  x        denote  the  number  of  shillings  at  first  ; 

Then  will  x  —     +  3       "  "  "       after  first  sitting  ; 

and 


will  denote  what  he  finally  had  ; 

hence,  from  the  conditions  of  the  problem, 


clearing  of  fractions,  performing  indicated  operations, 
and  reducing, 

6*  =  120  ;  .  •  .  x  =  20. 

29.  Two  persons,  A  and  B,  lay  out  equal  sums  of  money  in  trade  ; 
A  gains  $126,  and  B  losefc  $87,  and  A's  money  is  now  double  B's  : 
what  did  each  lay  out  ? 

Let  x  denote  the  number  of  dollars  laid  out  by  each  ; 

Then  will    x  +  126      "  "  "      A  had  ; 

mid  x  —  87        "  "  "      B     " 

From  the  conditions  of  the  problem, 

x  +  126  =  2  (x  —  87)  ; 

performing  indicated  operations,  transposing  and  reducing, 
—  x  =  —  300  ;  .  •  .  x  =  300. 

30.  A  person  goes  to  a  tavern  with  a  certain  sum  of  money  in  his 


4:2  KEY  TO  DAVIEB'  BOURDON.  [99-100. 

pocket,  where  he  spends  2  shillings;  he  then  borrows  as  much 
money  as  he  had  left,  and  going  to  another  tavern,  he  there 
spends  2  shillings  also ;  then  borrowing  again  as  much  money  as 
was  left,  he  went  to  a  third  tavern,  where,  likewise,  he  spent  2 
shillings  and  borrowed  as  much  as  he  had  left ;  and  again  spend- 
ing 2  shillings  at  a  fourth  tavern,  he  then  had  nothing  remaining. 
What  had  he  at  first? 

Let  x  denote  the  number  of  shillings  at  first. 
Then,  from  the  first  condition, 

x  —    2  will  denote  what  he  has  after  1st  visit. 

2(x-    2)-2or2z-    6         "  "  "         "    2d     '• 

2(2*—    C)  —  2or4«  —  14         "  "  «         "    3d     " 

2  (4r  -  14)  —  2  or  Sx  —  30         "  "  "         "    4th     " 

From  the  conditions  of  the  problem, 

8x  -  30  =  0     or     Sx  =  30 ;  .  • .  x  =  3f . 

or  the  amount  at  first  was  3s.  9d. 

31.  A  farmer  bought  a  basket  of  eggs,  and  offered  them  at  7  cents 
a  dozen.  But  before  he  had  sold  any,  5  dozen  were  broken  by  a 
careless  boy,  for  which  he  was  paid.  He  then  sold  the  remainder  at 
8  cents  a  dozen,  and  received  as  much  as  he  would  have  got  for  the 
whole  at  the  first  price.  Kow  many  eggs  had  he  in  his  basket  ? 

Let        x          denote  the  number  of  dozens  at  first ; 
Then  will  x,  -  5       "  «  "       sold ; 

and  8  (a;  —  5) "  cents  received ; 

7x  "  "  "     first  asked ; 

hence,  1x  —  8  (x  —  5)  ;  .  • .  x  =  40. 


100]  EQUATIONS    OF   THE    FIKST   DEGREE.  43 

32.  A  cask,  A,  contains  a  mixture  of  12  gallons  of  wine 
and  18  gallons  of  water;  another  cask,  B,  contains  a  mixture 
of  9  gallons  of  wine  and  3  gallons  of  water :  how  many  gal- 
lons must  be   drawn  from  each  to  produce  a  mixture   of  7 
gallons  of  wine  and  7  gallons  of  water  ? 

Let  x  denote  the  number  of  gallons  drawn  from  the  cask 
A,  and  14  —  x  the  number  of  gallons  drawn  from  the  cask  B. 

Of  the  x  gallons  drawn  from  A,  ^f  ths  is  wine  and  -^f  ths  is 
water;  in  like  manner,  of  the  mixture  drawn  from  B,  ^ths 
is  wine,  and  -^ths  is  water.  Hence,  all  the  wine  drawn 

12          9 

from  both  is  equal  to  ^ x  +  —  (14  —  x)  gallons;  but  this  is 
oU  1* 

equal  to  7  gallons.     Hence, 

12          9  , 

_*  +  _(14-*)  =  7, 

<*  "   ri   i  \  M 

or,  -*  +  -(U  — s)s7; 

x  =  10    and    14  —  x  =  4. 

33.  At  what  time  between  1  and  2  o'clock  is  the  minute 
hand  of  a  clock  just  1  minute  space  ahead  of  the  hour  hand  ? 

Let  x  denote  the  number  of  minute  spaces  passed  over  by 
the  hour  hand  from  12  o'clock  till  the  hands  have  the  re- 
quired position.  Then  61  +  x  will  denote  the  number  of 
minute  spaces  passed  over  by  the  minute  hand  in  the  same 
time;  but  the  minute  hand  travels  12  times  as  fast  as  the 
hour  hand.  Hence, 

12z  =  61  +  x, 
or,  llx  =  61 ; 

x=5JI    and    61+2  =  66^. 

That  is,  the  hands  will  have  the  required  position  at 
Ih.  G 


44  KEY   TO   DA.  VIES'    BOURDON.  [100. 

34.  A  person  having  a  hours  at  his  disposal,  how  far  can 
he  ride  in  a  coach  that  travels  0  miles  per  hour,  and  return 
on  foot  at  the  rate  of  c  miles  per  hour? 

Let  x  denote  the  number  of  miles. 

The  time  required  to  ride  x  miles  in  the  coach  will  be  de- 

/>» 

noted  by  j-,  and  the  time  required  to  walk  back  will  be  de- 

/v* 

noted  by  -.    From  the  conditions  of  the  problem,  we  have, 
c 

XX  „ 

T  -f-  -  =  a.    Hence, 
0      c 

(0  +  c}x  =  abc\    .'.    x  =  j-  --  . 

6  +  c  J 

35.  A  can  do  a  piece  of  work  in  one-half  the  time  that  B 
can  ;   and  B  can  do  it  in  two-thirds   the  time   that  C   can  ; 
all  together  can  do  it  in  6  days.     How  many  days  would  it 
take  each  to  do  it  singly? 

Let  x  denote  the  number  that  it  will  take  A  to  do  it. 
Then  will  %x  denote  the  number  of  days  that  it  will  take  B 
to  do  it  ;  and  3x  will  denote  the  number  of  days  it  will  take 

C  to  do  it.     Consequently,  A  can  do  a  part  denoted  by  - 

«v 

in  one  day,  B  can  do  a  part  denoted  by  ^-,   C  can  do  a 

/iX 

part  denoted  by  -$-  in  the  same  time,  and  all  together  can 

oX 

do  a  part  denoted  by  -  in  one  day.     Hence,  from  the  con- 
ditions of  the  problem, 


=       or 


.-.    -  =  -L    or,    a:  =  11,    2x  =  22,    and    3z  =  33. 
x      11 


106.]  EQUATIONS  OF  TIIE  FIRST  DEGREE. 

SIMULTANEOUS  EQUATIONS  OF  THE  FIRST  DEGREE. 

(2*4-  3y  =  16) 

1.  Given         •{  >  to  find  x  and  y. 

(3*  —  2y  =  11 ) 

Multiply  both  members  of  the  first  by  2,  and  of  the  second  by  3  j 


+  Gy  =  32 ) 


whence,  by  addition,  member  to  member,  we  have, 

13*  =  65  ;  .  • .          x  =  5,        also,        y  =  2. 


2.  Given 


2f      3y_  9_ 
"5" H"  T  ~  20" 

3s      2y_  61 
T  +  "5"  ~  120  J 


i.    to  find  x  and  y. 


Clearing  of  fractions,  and  then  multiplying  both  members  of  the  first 
by  16,  and  of  the  second  by  5, 


128*  +  240y  = 
450*  +  240y 

whence,  by  subtracting,  member  from  member. 


=  144) 
=  305) 


322*  =  161 ; 


3.  Given 


x  =  -,     also,  by  substitution,  y  =  -• 
*  3 


|  +  7*  =  51 


to  find  x  and  y. 


Multiplying  the  first  by  343  and  the  second  by  7 ; 

49*  f  2401y  -  33957  ) 
y  f-    49*    =    357    \ 


4r6      •  KEY  TO  DA  VIES*  BOURDON.  [106-107 

by  subtraction, 
24(%  =  33600 ;     .  • .     y  =  14  ;     also,  by  substitution,    x  =  7. 


4.  Given 


;  to  find  ar  and  y. 


5       '    3  4 

Clearing  of  fractions  and  transposing, 

2*  -  y  =  80 
47«-18y-2100; 

multiplying  both  members  of  the  first  by   18,  and  subtracting  the 
result  from  the  second,  member  from  member, 
11  x  —  660 ;         .  • .         x  =  60 ;         by  substitution,         y  =  40. 


5.  Given  * 


x  +    y  +    2  =  29 
*  +  2y  +  3z  =  62 


(1) 

(2) 

(3) 


*•  to  find  x,  y  and  *. 


Combining  (1)  and  (2), 

y  +  2z  =  33     ....     (4); 
combining  (1)  and  (3) 

2y  +  3z  =  54     .     •     •     •     (5) ; 
combining  (4)  and  (5) 

2  =  12;     by  successive  substitutions,   x  =  8.     y  =  9. 

pZ*  +  4y  -  3*  =  22     t     -     (I)' 

6.  Given  -I  4*  —  2y  -f  5z  =  18     •     •     (2)  [•;  to  find  r  -i  and  2. 
[&»  4-  ly  -    z  =  63  ( 

Combining  (1)  and  (2). 


107.]                      EQUATIONS   OF   THE  FIRST   DEGREE. 

10y-  112  =  26     .  .     .     (4); 

combining  (1)  and  (3), 

5y  -  Sz  =    3     .  .     .     (5)  ; 
combining  (4)  and  (5), 

5z  =  20  ;  . ' .         2  =  4. 

By  successive  substitutions,  x  =  3,         y  =  7. 


7.  Given 


•+I+I-* 


>.  ;  to  find  x,  y  and  ». 


7    i     eT    '     a  — 

.45       o 

Clearing  of  fractions, 

6x  +    3y  +    22  =  192     .     .     .     (1) ; 
20*  +  15y  +  122  =  900     .     .     .     (2) ; 
15x  +  12y  +  102  =  720     .     .     .     (3) ; 
combining  (1)  and  (2), 

16x  +  3y  =  252    .     .     .     (4) ; 
combining  (1)  and  (3), 

15ar  +  3y  =  240     .     .     .     (5); 

combining  (4)  and  (5), 

x  =  12; 

by  successive  substitutions,    y  =  20,     z  =  30. 

(  7z  -  22  +  3tt  =  17  .  .  .  (1)  " 

:y-2z+     t=l\  .  .  .  (2) 

3.  Given  ^   5y  —  3x  -  "2u  •—    8  .  .  .  (3)   I  ;  to  find  x,  y, 

\y  —  3?/  -f  2^  =    9  .  .  .  (4)         2,  t  and  «. 

3z  +  8w  =  33  .  .  .  (5) 


48  KEY   TO   DAVIES*   BOUEDON.  107. 

Combining  (2)  and  (4), 

4y  —  4z  +  3w  =  13     .     .     .     (6) ; 
combining  (1)  and  (3), 

35y  _  62  -  5w  =  107    .     .    .     (7)  ; 
combining  (5)  and  (6), 

12y  -1-  41w  =  171     .     .     .     (8); 
combining  (5)  and  (7), 

35y  +  ll«  =  173     .     .     .     (9); 
combining  (8)  and  (9), 

1303^  =  3909;  .-.         «=3; 

by  successive  substitutions,     a:  =  2,     y  =  4,     2  =  3,     /  =  1. 

C3x  +  2y  —  4:z  =  15 (1) 

9.  Given,  J  5x  —  3y  +  2z  =  28 (2) 

[  3y  +  ±z  —    x  =  24 (3) 

Combining  (1)  and  (3), 

ll#  +  8z  =  87 .,(4) 

Combining  (2)  and  (3), 

12y  +  22z  =  148.    ......    (5) 

Combining  (4)  and  (5), 

73y  =  365;    .-.    y  =  5. 
Substituting  and  reducing, 

*  =  7,    and    2  =  4. 


10.  Given, 


x      y 

M=» (») 

x      z 

-L     .     J.          o  ,-.» 


108.] 


EQUATIONS    OF    THE    FIRST    DEGREE. 


Adding  (1),  (2),  and  (3),  member  to  member,  and  divid- 
ing by  2, 

-      -      ---  (4) 

Subtracting  (1),  (2),  and  (3),  successively,  from  (4), 

J.          0  ^t 

2J  4  O 

i_i 

13  4 

-  +  -  =  - (1) 

x      y      z 

11.  Given, 

1      1_4 

Making    -  =  x',    -  =  y',    and    -  =  z't 
x  y  z 

2x'  +    y'  =  3z' (4) 

Bz'  —  2y'  =  2 (5) 

x'  +    z'  =  - (6) 

Combining  (4)  and  (5), 

4z'  +  3z'  =  Gz'  +  2, 
or,  4z'  —  3z'  =  2 (7) 

Combining  (6)  and  (7), 

,      6  7 

H  rp     — —    p\  •  •  />•     ^^   AT*  1*   — —  — 

4  "-t         —     \J    y  .        *  »t/       —     p,    J  Ul  ^  *O     — ^     p    • 

Substituting  and  reducing, 

,  _  10  _  21 

-21'    °r>        -10' 

2  7 

and,  y  =  —  = ,    or,    y  =  —  T. 


50 


KEY    TO    DAVIES'    BOUKDON.  [108. 

—  1  _  6z       x      9 

4~     =  1T~2  +  5  '     '    *    '* 

12.  Given,         l¥  +  ir  =  2/+!j (2) 

~^~  •*•  _  JL  I  1  —  _  I  y_  t<\\ 

7  14  "*"  6  ~~  21  "*"  3  ' 

Clearing  of  fractions  and  transposing, 

15y  +  10#  —  242  =  41 (4) 

—  12#  +  15x  +  lGz=  10 (5) 

—  14y  +  18z  —    7z  =  —  13 (6) 

Combining  (4)  and  (5), 

115z  —  162  =  214 (7) 

Combining  (4)  and  (6), 

410s  -  441z  =  379 (8) 

Combining  (7)  and  (8), 

8831z  =  8831 ; 

Substituting  and  reducing, 

x  =  2,      and     y  •=.  3. 

13.  Given,  ^  +  -  =  1 (2) 

c 


Adding  (1),  (2),  and  (3),  and  dividing  by  2, 
x      v      z       3 


108.] 


EQUATIONS    OF    THE    FIRST    DEGEEE. 


51 


Subtracting  (1),  (2),  and  (3),  successively,  from  (4), 
z       1  c 


a 

;;   .-.   *  =  -. 

7x-3y  =  I (1) 

llz-7w=l (2) 

14.  Given, 

4z  -  7y  =  1 (3) 

^  L9z  —  3w  =  l (4) 

Combining  (1)  and  (4), 

21w  —  57y  =  12 (5) 

Combining  (2)  and  (3), 

28w  —  77y  =  7 (6) 

Combining  (5)  and  (6), 

3^  =  27;    .-.    y  =  9. 
Substituting  and  reducing, 

x  —  4,    z  =  16,    and    u  =  25. 

T1 (1) 

15.  Given, 

Clearing  of  fractions  and  transposing, 

llx  -  My  =  -  30 (3) 

y  =  6,    and    x  =  12. 


52  KEY  TO  DAVIEB'  BOUEDON.  [108-111. 

^_l._  i  -  £._.£. 
10      15       9  "~  12      18 

- 

3~12      16      10 

Clearing  of  fractions,  transposing,  and  reducing, 

39z  —  2y  =  80 (3) 

U5x  +  ty  =  22G (4) 

Combining  (3)  and  (4), 

193z  =  386 ; 
x  =  %,    and    y  =  — 1. 

PROBLEMS  GIVING  RISE  TO   SIMULTANEOUS   EQUATIONS  OF 
THE  FIRST  DEGREE. 

5.  What  two  numbers  are  they,  whose  sum   is  33  and  whose 
iifference  is  7? 

Let  x  denote  the  first,  and  y  the  second. 
From  the  conditions, 

*  4-  y  —  33 

*-y=    7; 
whence,  by  combination, 

x  =  20,  y  =  13. 

t>.   Divide  the  number  75  into  two  such  parts,  that  thret  times  fa 
greater  may  exceed  seven  times  the  less  by  15. 

Let  x  dei.ote  the  greater,  and  y  the  less 
From  the  conditions  of  the  problem, 


Ill,    112.]         EQUATIONS   OF   THE  FIRST   DEGREE.  •    53 

x  -f-    y  =  75 

3x  -  7y  =  15  ; 

by  combination,     lOy  =  210  ;         .-.     y  =  21 ;     also,     x  =  54. 

7.  In  a  mixture  of  wine  and  cider,  ^  of  the  whole  plus  25  gallons 
was  wine,  and  %  part  minus  5  gallons,  was  cider  :  how  many  gallons 
were  there  of  each  ? 

Let        x  denote  the  number  of  gallons  of  wine  ; 
and  y         "  cider; 

then  will    x  -f  y  "  mixture. 

From  the  conditions, 

x  -4-  v 

— J-^  +  25  =  x 


3 

clearing  of  fractions,  transposing  and  reducing, 

y  —  x  =  —  50 
—  2y  +  x  —       15 
by  combination, 

y  =  35  ;         and         x  =  85 

8.  A  bill  of  £120  was  paid  in  guineas  and  moidores,  and  the 
number  of  pieces  of  both  >orts  that  were  used  was  just  100;  if  the 
guineas  were  estimated  at  21s.,  and  the  moidores  at  27s.,  how  many 
were  there  of  each  1 

Let     x  denote  the  number  of  moidores ; 
and         y  "  guineas ; 

then,  shce  £120  =  2400s.,  we  have,  from  the  conditions, 


54  KEY  TO  DAVIES'  BOURDON.  [112. 

x  +      y  =    100 
27*  +  21y  =  2400  ; 
by  combination,     %  =  300 ;     .  • .     y  =  50 ;     also,     x  =  50. 

9.  Two  travellers  set  out  at  the  same  time  from  London  and 
York,  whose  distance  apart  is  150  mile*  ;  they  travel  toward  each 
other;  one  of  them  goes  8  miles  a  day,  and  the  other  7;  in  what 
time  will  they  meet? 

Let         x  denote  the  number  of  miles  travelled  by  the  first ; 


V 

X 

then  will 

o 


and  y- 

From  the  conditions, 


days 


x  +  y  =  150 ; 

x__y_ 
8~7' 


second ; 
first; 

second ; 


whence,  by  combination, 
x  =  80        and 


Uio, 


the  number  of  days. 


10.  At  a  certain  election,  375  persons  voted  for  two  candidates ; 
and  the  candidate  chosen  had  a  majority  of  91 ;  how  many  voted  for 
each? 

Let  x  denote  the  number  of  votes  received  by  the  first ; 

y        "  "  "  "          "       second; 

from  the  conditions  of  the  problem, 

x  +  y  =  375 
x  =  y+    91; 
by  combination,  x  —  233,         y  =  142. 


112.]  EQUATIONS  OF  THE  FIRST  DEGKEE.  55 

11.  A's  age  is  double  B's,  and  B's  is  triple  C's,  and  the  sum  of 
all  their  ages  is  140  :  what  is  the  age  of  each  ? 

Let  x  denote  the  age  of  A  ; 

y  B; 

z         "  "  C; 

from  the  conditions  of  the  problem, 

*=    2y     •     •     •     •     (1); 
y=    3z     .     .     .     .     (2); 
x  +  y  +  z  =  140     .     •     •     •     (3)  ; 
from     (1)  and  ('2),  x  =  Gz  ; 

substituting,  y  =  3z,  and     x  =  62,    in  (3),  and  reducing, 

102  =  140 ;  .  • .  z  =  14,     x  =  84,     y  —  42. 

12.  A  person  bought  a  chaise,  horse  and  harness,  for  £60;  the 
norse  came  to  twice  the  price  of  the  harness,  and  the  chaise  to  twice 
the  price  of  the  horse  and  harness:  what  did  he  give  for  each? 

Let  x  denote  the  number  of  pounds  paid  for  the  harness; 
y         u  u  u  ((         «       horse; 

z         "  "  "  "         "       c'haise; 

from  the  conditions  of  the  problem, 

y  =  2x     •     •     •     •     (1) 
Z  =  2(x  +  y)     •     •     .     .     (2) 
x  +  y  +    z  =  GQ     •     •     •     •     (3) 
from  (2)     and     (1)     z  =  Gx; 

substituting         z  =  Gx      and      y  =  2x      in     (3) 
Jte  =  60,     .  ' .    x  =  6^,     also,  by  substitution,   y  =.  13£,     z  =  40; 

hence,  the  price  of  the  chaise  was  £40;  of  the  horse  £136*.  Sd. ; 
smd  that  of  the  hamss  £6  13s.  4d. 


56  KEY  TO  DAVIES'  BOURDON.  [112. 

13.  A  person  has  two  horses,  and  a  saddle  worth  £50  ;  now,  if 
the  saddle  be  put  on  the  back  of  the  first  horse,  it  will  make  his 
value  double  that  of  the  second  ;  but  if  it  be  put  on  the  back  of  the 
second,  it  will  make  his  value  triple  that  of  the  first  :  what  is  the 
value  of  each  horse  ? 

Let    x     denote  the  number  of  pounds  the  1st  horse  is  worth  j 

y         "  "  '          "     2d       "  " 

from  the  conditions  of  the  problem, 

x  +  50  =  2y 

y  +  50  =  3x  ; 
whence,  by  combination, 

x  =     30     y  =    40. 

14.  Two  persons,  A  and  B,  have  each  the  same  income.     A  saves 
£  of  his  yearly;  but  B,  by  spending  £50  per  annum  more  than  A, 
at  the  end  of  4  years  finds  himself  £100  in  debt  ;  what  is  the  income 
of  each  ? 

Let    x     denote  the  number  of  pounds  in  the  income  of  A  ; 
y         «  «  »  "...'«  B; 

by  the  conditions  of  the  problem,  these  are  equal  ;  one  only  will  be 

used.     Then  will 
4 

-x     denote  what  A  spends  per  year  : 
5 

-x  +  50"       "    B       "  " 

5 

» 

from  the  conditions  of  the  problem, 

4x+  100; 


whence,  performing  indicated  operations,  transposing  and  reducing, 
4<r  =    500  x  —     125. 


113.]  EQUATIONS  OF  THE  FIRST  DEGREE.  57 

15.  To  divide  the  number  36  into  three  such  parts,  that  \  of  the 
first,  £  of  the  second,  and  1  of  the  third,  may  be  all  equal  to  each 
other. 

Let  x,  y    and  0,  denote  the  parts. 
From  the  conditions  of  the  problem, 

x  +  y  +  x  —  36 

*_y 
2~3 

-  —  e 
2~45 

clearing  of  fractions,  and  combining, 

93  =  72     .*.     x  =  8  ;     whence,     y  =  12     and     2=16. 

16.  A  footman  agreed  to  serve  his  master  for  £8  a  year  and 
livery,  but  was  turned  away  at  the  end  of  7  months,  and  received 
only  £2  13s.  4d.  and  his  livery  :  what  was  its  value? 

Let  x  denote  the  value  of  livery,  expressed  in  shillings  :  £8  = 
160s.,  and  £2  13s.  4d.  = 


Then  will  I  —  rx  —  )  denete  the  value  of  wages  1  month, 


and 


by  the  conditions  of  the  problem, 


1120  +  7*  =  640  +  12  x 

—  5x=-  480 
x  =  96    .*.     value,  £4.16*. 


58  KEY  TO  DAVIKS'  BOURDON.  [113. 

17.  To  divide  the  number  90  into  four  such  parts,  that  if  the  first 
be  increased  by  2,  the  second  diminished  by  2,  the  third  multiplied 
by  2,  and  the  fourth  divided  by  2,  the  sum,  difference,  product 
and  quotient,  so  obtained,  will  be  all  equal  to  each  other. 

Let     x,  y,  z    and  u,  denote  the  parts  ; 
from  the  conditions  of  the  problem, 

x  +  y+z  +  u  =  QQ 

x+2=y  -2 
x  +  2  =  2z 


whence  we  find  from  the  last  three  equations, 

oc 
y  =  x  +  4,     2  —  -+1,     and     u  =  2x  +  4  ; 

ii 

substituting  these  values  in  the  first  equation, 

x  +  x  +  4  +  \  +  1  +  2ar  +  4  =  90  ;  or  4^r  =  81  ;  .  •  .    x  =  18  ; 

m 

whence,  by  substitution,     y  =  22,     z  =  10,     and     u  =  40. 

18.  The  hour  and  minute  hands  of  a  clock  are  exactly  together  at 
12  o'clock  :  when  are  they  next  together. 

1st  Solution. 

Let  x  denote  the  number  of  minute  spaces  passed  by  the  hour 
hand  before  they  come  together  ; 

and  y  the  number  passed  by  the  minute  hand  ; 

• 
then,   since  the  latter  travels  12  times  as  fast  as  the  former,  and 

since  it  has  to  gain  60  spaces,  we  have, 

*  —  y  =  60 


113.]  EQUATIONS  OF  THE  FIEST  DEGKEE.  59 

by  combination, 

lly  =  60  .  '  .  y  =  5T5T  ;      also,      x  =  65^  ; 

hence,  they  will  be  together,  65^-  minutes  after  12  o'clock,  or  at  4 
o'clock,  -fa  minutes,  and  at  the  end  of  every  succeeding  equal 
portion  of  time. 

2d  Solution. 

The  minute  hand  will  pass  the  hour  hand  11  times  before  they 
again  come  together  at  12  o'clock,  and  the  times  between  any  two 
consecutive  coincidences  will  be  equal.  Hence  each  time  will  be 
equal  to  12  hours  divided  by  11  =  l-^hr.  =  Ihr. 


19.  A  man  and  his  wife  usually  drank  out  a  cask  of  beer  in  12 
days  ;  but  when  the  man  was  from  home,  it  lasted  the  woman  30 
days  ;  how  many  days  would  the  man  be  in  drinking  it  alone  ? 

Let     x     denote  the  number  of  days  it  takes  the  man  to  drink  it  ; 
y         »  <c  u  u  woman  "         " 

then,  if  the  whole  quantity  of  beer  be  denoted  by  1, 

-  will  denote  the  quantity  drank  by  the  man  in  1  day  ;     and 
x 

-  "  "  "         "        woman  ; 

y 

from  the  conditions  of  the  problem, 

l  +  l-± 
x      y      12 


substituting  the  value  of  -  in  the  first  equation, 


- 

a?      30  ~  12  ' 


60  KEY  TO  DAVIES'  BOURDON.  [113. 

clearing  of  fractions, 

60  +  2*  =  5x  ;  .  •  .  x  =  20. 

»  20.  If  A  and  B  together  can  perform  a  piece  of  work  in  8  days, 
A  and  C  together  in  9  days,  and  B  and  C  in  10  days  :  how  many 
days  would  it  take  each  person  to  perform  the  same  work  alone  ? 

Let  the  work  be  denoted  by  1  ; 
Let     x     denote  the  work  done  by  A  in  one  day  ; 


then  will  -,  -   and  -  respectively  denote  the  number  of  days  that 
x  y  z 

It  will  take  A,  B,  and  C  severally  to  do  the  work  ; 
from  the  conditions  of  the  problem, 

x  +  y=    |     ....     (1) 

x  +  z=    I     •     .     .     .     (2) 

y  +  z=i5     •     .     .     .     (3); 

clearing  of  fractions, 

Sx+    Sy  =  l     •     •     •     •     (4.) 

9*  +    9z  =  1     .     •     •     •     (5) 

lOy  +  102  =  1     •     •     .     .     (6)  ; 

combining     (4)    and     (5), 

72y-  72*  =  1     •     •     •    •     (7), 
combining     (6)     and     (7), 


substituting  in     (1)     and     (3), 

X  =  ¥          7~2ff  —  TZO  >  Z  =  TO   ~~  Wo   =  T 


113-114.]  EQUATIONS  OF  THE  FIRST  DEGREE.  61 

21.  A  laborer  can  do  a  certain  work  expressed  by  a.  in  a  time 
expressed  by  b  ;  a  second  laborer,  the  work  c  in  a  time  </;  a  third, 
the  work  e  in  a  time  /.     Kequired  the  time  it  would  take  the  three 
laborers,  working  together,  to  perform  the  work  g  1 

If  a  laborer  can  do  a  piece  of  work  denoted  by  a,  in  a  number  of 
days  denoted  by  6,  he  can  do  in  1  day  so  much  of  the  work  as  is 

denoted  by  -r ;  the  second  in  1  day  can  do  so  much  as  is  denoted  by 

- ;  and  the  third  s<5  much  as  is  denoted  by  - ;'   hence,   the   three 
d  f 

working  together  can  do 

a       c       e  _  adf  -f-  bcf  +  bde 
b+d+f=         ~~bdf~ 

Let  x  denote  the  time  required  to  perform  the  work  g ;    then, 

the  three  can  perform  the  work  -  in  the  time  1 ; 

x 

from  the  conditions  of  the  problem, 

g  __  adf  -f-  bcf  +  bde 
~x~          ~~bdf 

taking  the  reciprocals  of  each  member,  and  then  clearing  of  fractions, 

we  have, 

bdfg 
~  adf  +  bcf  +  bde 

In  this  example  only  a  single  unknown  quantity  has  been  used, 
{gid  it  may  be  remarked  that  many  other  examples,  in  this  chapter, 
may  be  more  easily  solved  by  a  single  unknown  quantity  ;  in  such 
cases  more  than  one  has  been  used  for  the  purpose  of  illustration. 

22.  If  32  pounds  of  sea  water  contain  1  pound  of  salt,  how  much 
fresh  water  must  be  added  to  these  32  pounds,  in   order  that  the 


62  KEY  TO  DA  VIES5  BOURDON.  [114. 

quantity  of  salt  contained  in  32  pounds  of  the  new  mixture  shall  be 
reduced  to  2  ounces,  or  i  of  a  pound  ? 

Let  x  denote  the  number  of  pounds  to  be  added  ; 

then  will denote  the  number  of  pounds  of  salt   in  each 

32  -\-  x 

pound  of  the  mixture,  but  this  we  know  to  be  —  X  ^,  or,  — —  ; 

hence   from  the  conditions  of  the  problem, 

1  1 

= ,     or,     32  +  x  =  256,     or.     x  —  224. 

32  +  x       256' 

This  problem  is  also  solved  by  a  single  unknown  quantity  more 
readily  than  by  two. 

23.  A  number  is  expressed  by  three  figures ;  the  sum  of  these 
figures  is  1 1  ;  the  figure  in  the  place  of  units  is  double  that  in  the 
place  of  hundreds  ;  and  when  297  is  added  to  this  number,  the  sum 
obtained  is  expressed  by  the  figures  of  this  number  reversed. 
What  is  the  number  ? 

Let  #,  y    and  z  denote  the  digits  in  their  order ; 
then  will  the  number  be  denoted  by 

lOOar  -f  Wy+  z; 
from  the  conditions  of  the  problem, 

x+      y  +  z  =  ll     -     • (1) 

z  =    2x (2) 

100*  +  lOy  +  z  +  297  =  1000  +  lOy  +  *    •     •     •     (3); 
reducing  (3),  gives 

99*  -99*  :=297     •     •' (4); 

substituting  ?  =  2x  in  (4),  and  reducing, 

99.T  =  297  ;  . ' .  x  =  3  ; 


114.]  EQUATIONS  OF  THE  FIRST  DEGREE.  63 

whence,  by  successive  substitutions, 

y  =  2,         z  =  6.  Ans.  326. 

24.  A  person  who  possessed  $100000  dollars,  placed  the  greate? 
part  of  it  out  at  5  per  cent,  interest,  and  the  other  part  at  4  pei 
cent.  The  interest  which  he  received  for  the  whole  amounted  t< 
4640  dollars.  Required  the  two  parts. 

Let     x     denote  the  greater  part  ; 

y         "         "  lesser       " 
From  the  conditions  of  the  problem, 

5z 

-—  =  interest  on  x  dollars  at  5  per  cent.  ; 

^jfQ=         "          y        «          4    «       «       then, 
x  4-  y  =  100000     ....     (1) 


clearing  (2)  of  fractions, 

5*  +  4y  =  464000     •     .     •     •     (3) 
combining  (1)     and     (3), 

y  =  36000,         whence         x  =  64000. 

25.  A  person  possessed  a  certain  capital,  which  he  placed  out  at 
a  certain  interest.  Another  person  possessed  10000  dollars  more 
than  the  first,  and  putting  out  his  capital  1  per  cent,  more  ad  van- 
tageously,  had  an  income  greater  by  800  dollars.  A  third,  possessed 
15000  dollars  more  than  the  first,  and  putting  out  his  capital  2  per 
cent,  more  advantageously,  had  an  income  greater  by  1500  dollars. 
Required  the  capitals  and  the  three  rates  of  interest. 


64  KEY  TO  DAVIES'  BOURDON.  [114. 

Let  x  denote  the  number  of  dollars  in  1st  capital  ; 
and      y  the  rate  per  cent.  ;  then, 

y  X  y'   will  denote  the  number  of  dollars  of  1st  income  : 
100 

(y  +  10000)  (y  -*  !)_         „  „  M         „    2d 

100 

(x  f  15000)  (y  +  2)  u         u  u 

"155"" 

from  tho  conditions  of  the  problem, 

y 
"  " 


IOO  00 

15000)  (y  +  2)_ 


" 


100  oo 

cka.-'ng  of  fractions,  performing  indicated  operations,  transposing 
and  reducing, 

lOOOOy  +    x  =    70000 

15000y  +  2*  =  120000  ; 
combining  and  reducing,* 

5000y  =  20000        .  •  .        y  =  4  ;         and 
Dy  substitution,  x  =  30000  ; 

2d.  $40000,     rate  5  per  cent. 
3d.  $45000,     rate  6     "       " 

26.  A  cistern  may  be  filled  by  three  pipes,  A,  B,  C.  By  the  two 
first  it  can  be  filled  in  70  minutes  ;  by  the  first  and  third  it  can  be 
filled  in  84  minutes  ;  and  by  the  second  and  third  in  140  minutes. 
What  time  will  each  pipe  take  to  do  it  in?  What  time  will  be 
required,  if  the  three  pipes  run  together  ? 

Call  the  contents  of  the  cistern  1  , 


115.]  EQUATIONS  OF  THE  FIRST  DEGREE.  65 

Let  x  denote  the  quantity  discharged  in  1  minute  by  the  first  ; 
y         "  '  "  "         "         "       second; 

z  "  "  "         "         "       third; 

then  will    -,     -      and     -     denote  the  number  of  minutes  required 
*'    y  z 

for  the  pipes,  separately,  to  fill  the  cistern  ;        and, 

_1  _ 

'  *  +  y  +  z' 

will  denote  the  number  of  minutes  required  for  all  three  fc   fill  it, 

running  together  ; 

from  the  conditions  of  the  problem, 


•    •     •    •     (3) 


clearing  of  fractions, 

70*  +    70y  =  1    •     •     •     •     (4) 
84z  +    840  =  1    •     •     •     •     (5) 
140y  +  140z  =  1    •     •     •     •     (6)  ; 
combining  (1)  and  (2), 

840y  -  840z  =2    •    •    -     •     (7)  ; 
combining  (6)  and  (7), 


substituting  in  (1),  and  transposing, 


JL    _L     A    JL. 

*  ~  70      210  ~  210  ~  105  ' 


66  KEY  TO  DA  VIES'  BOURDON.  [115. 

substituting  in     (3), 

1  1  1 

~  140  ~  210  ~~  420' 

*  +  y  +  *=:2To+l05  +  420  =  601 

hence,      -  =  105.      -  =  210       -  =  420.     -1 =  60. 

x  y  z  x-|-y-|-2 

27.  A  has  3  purses,  each  containing  a  certain  sum  of  money.  If 
$20  be  taken  out  of  the  first  and  put  into  the  second,  it  will  contain 
four  times  as  much  as  remains  in  the  first.  If  $60  be  taken  from  the 
second  and  put  into  the  third,  then  this  will  contain  If  times  as 
inuch  as  there  remains  in  the  second.  Again,  if  $40  be  taken  from 
the  third  and  put  into  the  first,  then  the  third  will  contain  2£  times 
as  much  as  the  first.  What  were  the  contents  of  each  purse  ? 

Let     x     denote  the  number  of  dollars  in  the  first  purse. 
y         "  "  "          "      second  " 

z         "  "  "          "      third      " 

then  from  the  conditions  of  the  problem, 
4  (x  -  20)  =  y  +  20, 
I  (y  -  60)  =  z  +  60, 

2_40   -2*.(x  +  4ff)-, 

clearing  of  fractions,  performing  operations  and  transposing, 
4x  -      y=    100     ....     (1) 
7y  -     4z  =    660          .     .          (2) 
8z  -  23.r  r=  1240     ...          (3); 
combining     (1)     and     (2), 

28z-4z7=  1360    ....     (4); 


115.]  EQUATIONS  OF  THE  FIRST  DEGREE.  67 

combining     (3)     and     (4), 

33*  =  3960;     .'.     x  =  120  ; 

by  substitution,  y  =  380;     z  =  500. 

28.  A  banker  has  two  kinds  of  money ;  it  takes   a  pieces  of  the 
first  to  make  a  crown,  and   b   of  the  second  to  make  the  same  sum. 
Some  one  offers  him  a  crown  for   c  pieces.     How  many  of  each 
kind  must  the  banker  give  him  1 

Since  it  takes   a   pieces  of  the  first  to  make  1  crown,  -  —  thepai't 
of  a  crown  in  each  piece ;  and   -,    the   part  of  a  crown   in   each 

piece  of  the  second : 

let     x     denote  the  number  of  pieces  taken  of  the  first  kind, 
y  "  "  "  "  second  " 

from  the  conditions  of  the  problem, 

x  +  y  —c 

x       y 

-  +  j-  =  1,     or     bx  -f  ay  =,ab  ; 

Cl          0 

by  combination, 

by  —  ay  =  be  —  ab  ;     or     y  (b  —  a)  =  b  (c  —  a) ; 

(a  —  c)  b  a  (c  —  b) 

.'.     y  =  - r— ;      whence,     x  =  — ^. 

a  —  b  a  —  b 

29.  Find  what  each  of  three  persons,  A,  B,  C,  is  worth,  knowing 
1st,  that  what  A  is  worth  added  to  I  times  what  B  and  C  are  worth 
is  equal  to  p ;  2d,  that  what  B  is  worth  added  to  m  times  what  A 
and  C  are  worth,  is  equal  to  g;    3d,  that  what  C  is  worth  added  tc 
n  times  what  A  and  B  are  worth,  is  equal  to  r. 


68  KEY  TO  DAVIE8*  BOURDON.  [115 

Let    x     denote  what  A  is  worth, 
y         «:         «      B         " 

2  «  «         C  " 

then,  from  the  expressed  conditions. 

x  +  l  (y  +  z)  =P  •  »  .  .  (1) 
y  +  m  (x  +  z)  =  q  •  .  .  .  (2) 
2  4-  n  (x  +  y)  =  r  -  •  .  .  (3)  ; 

waich,  by  adding  and  subtracting  Ix,  my   and  KZ,  may  be  written 
under  the  forms 


z)=p  ....  (4) 
(1  —  m)  y  -f  m  (x  -f-  y  +  z)  =  q  •  •  -  -  (5) 
(l-n)z+n(z  +  y  +  z)  =  r  -  •  •  •  (6)  ; 

dividing  both  members  of  each  equation  by  the  co-efficient  of  its 
first  term, 


—  m  1  —  m 

n  r 


(8) 


adding  these,  member  to  member,  and  deducing  from  the  resulting 
equation  the  value  of  x  +  y  +  2, 


1  _  (  '    \-m  '    l-n 


1  -  l       1  -  tfi       1  —  n 

Denote  the  second  membei  of  equation   (10)  by  the  single  letter  *, 


115.]  EQUATIONS  OF  THE  FIRST  DEGREE.  69 

a  known  quantity.  Then  by  substituting  this  for  the  factor 
a  4-  y  -f  z,  in  each  of  the  equations  (7),  (8)  and  (9),  and  dedue- 
>'.g  the  values  of  x,  y  and  z,  we  have, 


~ 


v=  ms 

" 


1  —  m       1  —  m        I  —  m 

r  ns  r  —  ns 


1  —  n        1  —  n        1  —  » 


(13). 


Had  we  represented  the  polynomial  x  +  y  +  z  by  s,  the  alge- 
braic work  would  be  slightly  diminished,  but  the  preceding  method 
has  been  followed  in  order  to  show  more  clearly  the  process  of  solu- 
tion. 

30.  Find  the  values  of  the  estates  of  six  persons,  A,  B,  C,  D,  E, 
F,  from  the  following  conditions  :  1st.  The  sum  of  the  estates  of  A 
and  B  is  equal  to  a  ;  that  of  C  and  D  is  equal  to  b  ;  and  that  of  E 
and  F  is  equal  to  c.  2d.  The  estate  of  A  is  worth  m  times  that  of 
C ;  the  estate  of  D  is  worth  n  times  that  of  E,  and  the  estate  of  F 
is  worth  p  times  that  of  B. 

1st  Solution. 

Let    x    denote  the  value  of  A's  estate ; 
a  —  x  "  "      "   B's       " 

y  •'  "       ;<    C's        " 

b  —  y  "  "      "   D's      " 

z         "  "      "   E's      '* 

c  _  g    «•  «        «     F'S         « 

from  the  conditions  of  the  problem, 


70  KEY  TO  DAVIES'  BOURDON.  [115. 


*  =  my 

b  —  y  =  nz 

c  —  z  =  p  (a  —  x) 


or, 


x  —  my  =  0 (1) 

y  +  nz  =  b  •     .     •          •     (2) 
px  —  z  =  ap  —  c     •     •          (3) 


combining  (1)  and  (2), 

x  -f-  muz  =  bm     ....     (4) ; 
combining  (3)  and  (4),  and  finding  the  value  of  2, 

bmp  +  c  —  ap 

z  =  — ; •  ,      .  which  denote  by  P : 

mnp  +  1 

combining  (3)  and  (4),  and  finding  the  value  of  a?, 

amnp  +  bm  —  cmn 

x  = — ,  which  denote  by  O ; 

mnp  H-  1 

substituting  the  last  value  in  (1),  and  finding  the  value  of  y, 

anp  +  b  —  en      ,.,,       ,,       D 

y  =  — * —  which  denote  by  R : 

mnp  +  1 

whence, 

A's  estate  is  equal  to        Q, 

B's     "         "         "     a  —  Q, 

C's     "        «        "  R, 

D's    "        «        "     b  —  R, 

E's    «        "        "  P, 

F-s    "        «        "     c  -  P. 

We  might  have  combined  (2)  and  (3),  eliminating  2,  and  then  from 
this  resulting  equation,  taken  with  (1),  have  found  the  value  of  a 
and  y. 

2d.  Solution 

By  a  single  unknown  quantity, 

Let  x  denote  the  value  of  C's  estate ; 

'hen  will  b  -  x  "  "          D's      « 


127.]  INEQUALITIES.  71 

mz  denote  the  value  of  A's       " 

a  —  mx  "  "          B's        " 

p(a-mx)         "  "  F's       " 

c  —p(a-mx)"  "  E's       « 

and  from  the  remaining  condition    of  the  problem, 
b  —  x  =  n[c  —  p(a  —  mx)]  • 

whence,  by  the  rule  for  solving  equations  of  the  first  degree, 

b  -f  <wp  —  nc 

x  =. • 

mnp  -\-  1 

Having  found  the  value  of  C's  estate,  the  remaining  quantities 
may  be  found  by  substituting  it  in  the  expressions  of  the  data,  and 
reducing.  The  operations  are  obvious. 

INEQUALITIES. 

1.  Given,         5x  —  6  >  19,         to  find  the  smallest  limit  of  x. 
If  we  add  6  to  both  numbers  of  the  inequality,  we  have 

5z  >  19  +  6,     or,     5z>  25; 
dividing  both  numbers  by  5,  we  have 

ar>5. 

14 

2.  Given,         3x  +  —  x  —  30  >  10,     to  find  the  least  limit  of*. 

m 

Reducing, 3x+1x-  30  >  10,     or,     10z  —  30  >  10 ; 

adding  30  to  both  members  of  the  inequality,  and  dividing  by  10, 
sre  have, 

z>4. 

3.  Given,         J  —  -  +  -  f  -^  >  — ,  to  find  the  least  'imit  of  ST. 

D  O  <6  <*  '- 


72  KEY  TO  DAVIES    BOURDON.  [127-149. 

Multiplying  both  members  by  6,  we  have, 

x  —  2x  -f  3z  +  39  >  51  ; 

reducing,  subtracting  39  from  both  members,  and  dividing  by  2, 
we  have, 

x  >6. 

ax  cfl 

4.  Given,         -—  +  bx  —  ab  >  —  to  find  the  least  limit  of  x. 

5  5 

Multiplying  both  members  by  5,  we  have 

ax  -f  5fce  —  Sab  >  a2  ; 

adding  -f  5ab  to  both  members,  and  dividing  by  the  co-  efficients  of 
x,  we  have, 

(a  +  56)  x  >  a  (a  +  56)  ;     or,     x  >  a. 

&C  J2 

5.  Given,         —  —  ax  +  a&  <  —  ,  to  find  the  largest  limit  of  x. 

Multiplying  both  members  by  7,  adding  —  7a&,  and  dividing  by 
the  co-efficients  of  x,  we  have, 

(b  —  7a)  x  <  b  (b  —  7a)     or,     a  <  i. 

REDUCTION  OP  RADICALS. 

7 

1.  -  —.    Multiply  both  terms  by    3  +  \f5. 
3  —  A/5 

•y  /K 

2.  JlV  —  7=.    Multiply  ooth  terms  by    VH  —  V3. 


3.  Multiply  both  terms  by    5\/l3  +  6\/5. 

5A/I3-6-N/5 


150.]  REDUCTION    OF    RADICALS.  73 


-t  ^  terms 

(5  -  A/5)  (1  +  A/3) 


5  -{-  A/5    and    1  -  A/3;    this  S17 

(3  +  A/3)  (1  -  A/3)  (3  +  A/5)  (5  +  A/5)  (A/5  -2). 
(5  _  V5)  (5  +  A/5)  (1  +  A/3)  (1  -  A/3) 

which,  after  performing  the  operations  indicated,  and  reducing, 
becomes, 

-  40  A/15  —  80  A/3  +  80  \/3  +  32  A/15  _  1    rr= 
20z  —  2  ~5V 


—          ,r  ,,  .  ,  .  /  - 

5.  -  -     •          Multiplying  both  terms  by   \a-\-x 

V  a  +  x  —  A/a  —  a; 
A/«  —  ^j  we  have, 

2a  +  2  Aa2  — 


(7  -2  A/5)  +  2  +  A/45 

6.  -i  -    _1     —  =  -  .      Multiplying    both    terms   by 
2  v  3  —  A/7 

2  A/3  +  A/7,  the  denominator  becomes  5.    For  the  numerator, 
performing  the  multiplications  and  reductions,  we  have, 

7  _  2  V5~~+  2  +  A/45 
2A/3+  A/7 

U  ^3  —  4  A/15  +  4  A/3  +  2  A/135 

+  7  A/7  —  2  A/35  +  2  A/7  +  A/315. 

But,      2  A/135  =  6  A/15;    and    A/315  =  3A/35; 
hence,  the  product  reduces  to 

18A/3  +  2A/15+  A/35  +  9A/7. 


74  KEY    TO    DAVIES'    BOURDON.  [150-158. 

But, 

18  A/3  =  9  x  2  \/3;    and    2  <v/15  =  2  \/3  x  \/5  ; 
also, 

V35  =  V7  x  V55 
hence,  the  sum  reduces  to 


=  9  (2  \/3  +  V7)  +  A/5  (2  A/3  +  A/7) 
=  (9  +  A/5)(2A/3 


7.  Given         — -x  =  1 x — ,     to  find  the  values  of  x. 

3        o  a  3 

Clearing  of  fractions,  transposing  and  reducing, 

«2  —  M 


whence,  by  the  rule 


_  a2  —  bz  /        a*  _ 

~2^6~    b  V  ~~  ~"~ 


—  2a262  +  64  _  a2  —  bz       a2  +  62  . 
~~  ~"~          2ab 


2a2       a 
taking  the  upper  sign,          x  =  —  -  =  -, 


2bz  b 

lower     "       x  —  —  ---.  = 

2ab  a 


dx      3xz   ,  1  +  c      xz    t   x 

8  Given        —  +  —  +  1  = —  +  -v 

c          4  c  4       a 

to  find  the  values  of  x. 

Clearing  of  fractions,  transposing  and  reducing, 

d*-c         1 


159.]  EQUATIONS  OF  THE    SECOND   DEGREE.  75 

whence,  by  the  rule, 


\/7+- 


~  C 

— ^- 


2cd 


2c        1 

taking  the  upper  sign,          x  —  —  =  -., 

2d2  d 

lower     "        x  =  —  ^—,  =  --  • 
2cd  c 

xz      2x   ,   59  x2       x  . 

9.   Given     —  ---  —  +  —  =  8  —  -  —  -»    to  find  the  values  of  T. 
4         o         o  4        o 

Clearing  of  fractions,  transposing  and  reducing, 


whence,  by  the  rule, 

'"5~T~T      1       7 


3  5 

hence,  x  =  -      and     x  =  —  -• 

«  o 

A  -.  90         90  27 

10.  Given T— =•  =  — — ^1    to  find  the  -alues  of  x. 

x       x+  1       a;  +  2 

Dividing  both  members  by  9, 

10          10  3 


x       x  +  1        r  +  i* 

clearing  of  fractions,  transposing  and  reducing, 


20 


vrhence,  by  the  rule, 


76  KEY  TO  DAVIES'  BOFKDON.  [159. 

20 49~7       17 


10  5 

«  =  4,     and     x=--=~-    . 

11.  Given         — 2  = — ,  to  find  the  valuei  of  *. 

8  —  x  x  —  2 

Clearing  of  fractions,  &c., 

39  28 


r*£i    ^         />t    —  —     _         ^..     • 

5  5  > 


whence  by  the  rule, 


_39          /     28 

~  10      V  "r  5       "r7T7r  *~  n~ 


1521  _  39      31. 
100~~10       10' 


v,  j  8        4 

hence,  a;  =  7,     and-   a;  =  —  =  -. 

10       o 


12.  Given        ax —  -f-  b  = — - —  a;2  +  -a?,  to  find  the  values  of  a;. 

o  b  a 

Clearing  of  fractions,  &c., 

/*»2   _, *T«  —  A  • 

ti,  tt/      —      I/     . 

a 
whence,  by  the  rule, 


a2  -  6          A       a*  -  2a25  +  62      a2  -  i      a2  + 


:VA^ 


2a       "V                    4a2  2a  2a 

hence,  a;  =  a,         2;  = 

8 

a—  n  f%/)**          *7*          n     I     /i  o**^          r>2 

—  y       .    *>*          «          u  -f-  a  x          c; 

1  "-J    C-iivAn  -  /*•    I     ^-  _r  .__  .        —  „       />•     i 

c  2         c2~      c          H2       <*' 

to  find  the  values  of  x. 


159.]  EQUATIONS  OF  THE  SECOND  DEGREE.  77 

Clearing  of  fractions,  &c., 


26         a2  —  62 

ffi     __          „_  /)•    —    _^^______™ 

- 


whence,  by  the  rule, 


_&  A^= 

~C  V          C2 


4-    1  -  - 

~~   *"  ^  —  c        c 


b  +  a  b  —a 

hemce,  x  =  -  ,     and    x  = 


c  c 


14.  Given        mx2  +  mn  =  2m-y/n  •  x  +  nxz,  to  find  the  values  of  x. 
Transposing  and  reducing, 


xz 


m  —  n  m  —  n 

whence,  by  the  rule, 


m-\fn  I        mn  mzn  m-Jn        n-\fm 

—  _  y         -4—  A    /  ^_   _  _i_  --  —          T         -^»         *  _  . 

~  m  —  n      V       m  —  n       (m  —  n2       m  —  n      m  —  n  ' 


(m  —  n) 

m/n       n-vw 
aence,  z  = 


~Vn 

since,         m  —  n  =  (-y/m  +  -y/n)  (-y/w  —  -yA)^        (-^ 
also, 

m  y^T—  ny'm  _  \/rnn(\/m—  -y/n) 


V" 


2rf  Solution. 

mxz  -f-  wiw  =  2m  -y/n"-  ar  +  na;2  j 
transposing  the  first  term  of  the  second  member,  we  have, 


78  KEY  TO  DAVIES'  BOURDON.  [159. 

ma2  —  2m  -\fn  •  x  +  mn  =  nx2  ; 
observing  that  the  first  member  is  the  square  of  a  binomial  whose 

terms  are     -y/m  •  x  —  ^/m  y^   we  have, 

y#T  •  x  —  -y/m  •y/n  =  ±  y"n  •  x  ; 
-y/m  -  x  q=  -\/n  '•  x  —  -y/m  -y/jT 

-y/n)  x  =  y'my'n  :  hence, 


•\/mn  \/nm 

—  —  ^  --  -     and,    x  =  —L  --  —  • 
-y/m  —  y  n  ym  +  yn 


6a2       b*x       ab  - 
1  5.  Given         ato2  --  -  -)  --  = 


c2         c  c2 

<o  find  the  values  of  x. 
Clearing  of  fractions,  &c., 

ft2  +  3a2         aft  -  252  +  6a2 


whence,  by  the  rule, 


52  -4-  3a2  I  oh  —  262  -4-  6a2 

= ! ±  \  / ' \- 

2abc  V 


4a262c2 


3a2       3o2 


2abc  2abc 

4ab  —  2bz      2o  —  b  6a2  +  4ab  3a  -f 

hence,   a?  =  —  —  ^  --  =  -  ,    a;  =  --  - 


2a6c  ac  2aoc  oc 


4x2   ,2x  3jrz       58a: 

16.  Given        —  +  y-f  10  =  19  -  j-  +  -y-, 

find  the  values  of  x. 


159.]  EQUATIONS  OF  THE  SECOND  DEGREE.  79 

Clearing  of  fractions,  &c., 

x*  -  Sx  =  9  ; 
whence,  by  the  rule, 

x  =  4  ±  v/9  +  Iti  =  4  ±  5, 
hence,  a;  —  9,     a;  =  —  1. 

I 

17.  Given        6  = ,    to  find  the  values  of  x. 

x  —  a  a  +  x 

Clearing  of  fractions,  &c., 


6-2 
whence,  by  extracting  the  square  root  of  both  members, 


and   x=   -°- 


18.  Given        2z  -f  2  =  24  —  5x  —  2z2,  to  find  the  \alues  of  * 
Transposing  and  reducing, 

7 

whence,  by  the  rule, 


whence,  x  =  2,     and     a:  = — • 

SB 

19.  Given        x*  —  x  —  40  =  170,     to  find  the  values  of  «. 
Transposing, 


80  KEY  TO  DAVIES'  BOURDON.  [159. 

whence,  by  the  rule, 


hence,  x  =  15,     and    x  =  —  14. 

20.  Given        3z2  +  2z  —  9  =  76,     to  find  the  values  of  *. 
Transposing  and  reducing, 


whence,  by  the  rule, 


1          /JS.l  1  +  1^. 

8       VlF      9"~        33' 


17  2 

hence,  «  =  5,     and     x= --=  —  5  -• 

o  o 


fni^"3j 

21.  Given         a?  +  b2  —  2bx  +  z2  =  -  —  ,  to  find  the  values  ol  x 

ri2 

Clearing  of  fractions,  &c., 

ffi         \ 

i 


^^^________^^^ 

9  9*  —         5  5  —  ) 

m2  —  nz          m2  —  n2 
whence,  by  the  rule, 


bn2  /n2a2  +  n2!)2 

m2  — n2       V     m2  —  n2          m*  — 


2m2n2  -f  n« 


bn2  n  /  a2w,2  +  h2m2  —  a2n2 


w2  —  m2  n2  —  m2 


whence,         x  =  ^^^  \  bn  ±  /a2™2  +  62m2  —  a2w2  [ 


160.]  EQUATIONS    OF    THE    SECOND    DEGREE.  81 

22.  Given  _L__  +  _|_=1. 

whence,  2  (x  +  1)  +  12  =  z2  —  1, 

and  .'.     a;  =  l±yT6,    a;' =  5,    »"  =  — 3. 

a;  40  3  (10  -f  a;) 

23'GlVen          15 +  300^)=      4)5-    > 
whence,    19z  (10  —  x)  +  3800  =  9  (10  +  z)  (10  —  «). 

Reducing,  z2  —  19a;  =  290 ; 

19  ,  ,/i  ;~36i    19  ±39 

hence,  »  =  y±y290  +  —  =  — i — 

.-.    a;'=29,    a;"=-10. 

24.  Given  ^=^  =  »  -3  +  -; 

a;  +  <>  * 

whence,  x  (x*  —  5x)  =  (a?  —  9)a?  +  *  +  3. 

Reducing, 

hence, 


or 


25.  Given 


whence,  2s2  =  3  (x2  —  x)  +  2  (z2  —  2z  +  1) ; 

72  7  ,    .749      2 

hence,    o^--^--,       or       ^-ij/---; 

.-.    a;' =  2        and        «"  =  -. 

26   Given  g  +  2      s-2_5. 

a;_2      s  +  a-8' 

whence,    6  (a?  +  4x  +  4)  —  6  (z3  —  4a;  +  4)  =  5  (&  —  4); 

48 
therefore,  a?  — -  a;  =  4 ; 


KEY   TO  DAVIES'   BOUEDON. 


,  ,     . ,        ,  24  ,  ,/576~~       24±26 

and  hy  the  rule,    x=—±y  -^+4= — ;— — ; 

5  iiiD  O 

.*.    a;' =  10        and        s"=— f. 

5 

~x                    x  —  6       a;  — 12      5 
27.  Given  --: —  =  - ; 


whence,     6(z2—  12z+36)  —  6(z2—  24z  +  144)  =5(z2—  18a,-+  72) 

6 


162  1008 

.Reducing,  a;2  --  —  x  =  --  —  ; 


_  81       ./6561       5040  _  81  ±39 
=       ±1/         "  "       :       ~~ 


oo    n- 
28.  Given 


25 

'  =  24        and        a?"  =  ^ 
5 

a;          »  +  1       13 


x  +  l^      x  6 ' 

whence,         6a;2  +  6  (a?  +  2z  +  1)  =  13  (a;2  +  x) ; 

1          1 +5 

hence,  x  =  —  „  ±  v  6^  =  — 5 — ; 

/&  /* 

.-.    »'=  2        and        &"=  —  3. 

1  23 

29.  Given  — — -„  =  ^ ; 

whence,          5  (»  +  2)  —  10  (x  —  2)  =  3  (x*  —  4) ; 

5    ,  4/~  ~25           5  ±23, 
hence,          «=  -  ^  ±  |/  14  +  ^  = g— ; 

.-.    a?'=3        and        x"=—  y. 

,   „.  4  5  12 

30.  Given  — —=•  +  ,    ,  0  =  „  ,   .. ; 


whence,  4(a;2+5z+6)  +  5(^+4a;+3)=12(a?+3a;+2); 

2±7. 


hence,  x  =  ^  ± 

.•.    x'=  3        and        x"=—  5. 


162.]  EQUATIONS   OF  THE   SECOND   DEGREE.  83 


i'ROBLEMS  GIVING  RISE  TO  EQUATIONS  OF  THE  SECOND  DEGREE 

4.  A  grazier  bought  as  many  sheep  as  cost  him  £60,  and  after 
reserving  15  out  of  the  number,  he  sold  the  remainder  for  £54,  and 
gained  2s.  a  head  on  those  he  sold  :  how  many  did  he  buy  1 

Let     x     denote  the  number  purchased  : 
and  x  —  15,     the  number  sold  ; 

then  will  denote  the  number  of  shillings  paid  for  1  sheep, 

9 

and  -  —      the  number  of  shillings  received  for  each. 

a:  —  15 

From  the  conditions  of  the  problem, 

1200  _     1080 
x     ~  z^l5  "      ' 

clearing  of  fractions,  &c., 

x3-  +  45*  =  9000  : 
whence,  by  the  rule, 


_  45       195 


22' 

hence.  x  =  75,     and     x  —  —  120, 

the  positive  value  only,  corresponds  to  the  required  solution. 

5.  A  merchant  bought  cloth  for  which  he  paid  £33  15*.,  which  he 
sold  again  at  £2  8*.  per  piece,  and  gained  by  the  bargain  as  much 
as  one  piece  cost  him  :  how  many  pieces  did  he  buy  ? 

Let    x     denote  the  number  of  pieces  purchased  : 

675 

then  will,  denote  he  number  of  shillings  paid  for  each, 


fi4  KEY  TO  DAVIKS'  BOURDON.  [163. 

and  48#  the  number  of  shillings  for  which  he  sold  the  whole. 
From  the  conditions  of  the  problem, 


48* -675=^- 
x 


then,  by  clearing  of  fractions,  &c., 
whence,  by  the  rule, 


225         225 

, , , /*•    -— .     _  i 

16  16 


_225^       / 

X~  32  "  V 


225       50625       225      255 
~T6~        1024    :  ~  "32"  "  "32" 


480 
using  the  positive  value  only,         x  =  •——  =  15. 

HI 

6.  What  number  is  that,  which  being  divided  by  the  product  of 
its  digits,  the  quotient  will  be  3  ;  and  if  18  be  added  to  it,  the  order 
of  its  digits  will  be  reversed  1 

Let    x    denote  the  first  digit, 
and,       y        "  second  " 

ehen  will  Wx  -f  y        denote  the  number. 

From  the  conditions  of  the  problem, 

10*  +  y  _  o 

—  o, 

xy 

10a:-f-y+  18  =  1 
whence,  by  reduction, 


finding  the  value  of  x   in  terms  Df   y  from  the  second,  and  sub- 
sti  tnting  in  the  first,  we  have. 


163.]  EQUATIONS   OF  THE   SECOND   DEGREE.  85 

10y  —  20  +  y  =  3y2  —  6y  ; 
whence,  by  transposing,  &c., 


17  ,        20 

-—  y2  =  -—  ;      and, 


by  the  rule, 


17          /      20  ,  289        ,   17. 
-6-*  V  ""a  +  S6=+1T 


taking  the  positive  sign,  y  =  4  ; 

whence,  a,  =  2,     and  the  number  is  24. 

7.  Find  a  number  such  that  if  you  subtract  it  from  10,  and  mul 
tiply  the  remainder  by  the  number  itself,  the  product  will  be  21. 

Let    x     denote  the  number  : 
from  the  conditions  of  the  problem, 

(10  -  *)  x  =  21  ;     or,    xz  —  10*  =  -  21  j 
by  the  rule, 


x  =  5  ±  y  -  21  +  25  =  5  db  2; 
whence,  x  =.  7,     and     x  =  3. 

8.  Two  persons,  A  and  B,  departed  from  different  places  at  the 
same  time,  and  travelled  towards  each  other.  On  meeting,  it  ap- 
peared that  A  had  travelled  18  miles  more  than  B;  and  that  A 
could  have  performed  B's  journey  in  15|-  days,  but  B  would  have 
been  28  days  in  performing  A's  journey.  How  far  did  each  travel  1 

Let      x        denote  the  number  of  miles  B  travelled  ; 
x  +  18        "  "  "    A        " 

"  «  "    A        "        in  one  day, 


KEY  TO  DAVIES'  BOTJKDON.  [163. 

— denote  the  number  of  miles  B  travelled  in  one  day  , 

_    _l_    I  Q 

"  "  days  A 


15} 


tx  -f  18\ 

V28-; 


B 


from  the  conditions  of  the  problem, 

x+  18  _        x_  x2  H-  36s  -f  324  _  x* 

~>T    ~*  ~28  ~I5 


clearing  of  fractions,  and  reducing, 

2       324     _  2916 

'^Ta      ~' 

By  the  rule, 


162    /2916   26244 
:    :t~~   ~~' 


162   216 


nence,  using  the  upper  sign, 

378 
*  =  -   =  54. 


9.  A  company  at  a  tavern  had  £8  15s.  to  pay  for  their  reckoning  ; 
but  before  the  bill  was  settled,  two  of  them  left  the  room,  and  then 
those  whc  remained  had  10s.  apiece  more  to  pay  than  before:  how 
many  were  there  in  the  company  1 

Let  x  denote  the  number  in  the  company. 


163.]  EQUATIONS  OF  THE  SFCOND  DEGREE.  87 

175 

Then      —   will  denote  the  number  of  shillings  each  should  pay  ; 


"    paid; 


from  the  conditions  of  the  problem, 

17C  -|7C 

I/O  I/O 

*  —  2         *~  '"       ' 
clearing  of  fractions, 

175*  —  175*  +  350  =  10*2  -  20*; 
whence,  *2  —  2*  =  35. 

By  the  rule, 


x  =  1  ±y36=  1  ±6; 

using  the  upper  sign,        x  =  7. 

10.  What  two  numbers  are  those  whose  difference  is  15,  and  o* 
which  the  cube  of  the  lesser  is  equal  to  half  their  product? 

Let       x      denote  the  smaller  number  ; 
then  will  x  +  15  "  greater         " 

from  the  conditions  of  the  problem, 

*3  =  \  (3.2  +  15*),     or,     x*  =  1  (x  +  15)  ; 


1        15 

x   — 

By  the  rule, 


whence,  xz  —  -x  =—  . 

z         « 


4-a-V~2~^16-4-I-T 

using  the  upper  sign,         x  =  3  ;         hence,         *  -f  15  =  18. 

11,  Two  partners,  A  and  B,  gained  $140  in  trade:  A's  money 


88  KEY  TO  DAVIES*  BOUKDON.  [163. 

was  3  months  in  trade,  and  his  gain  was  $60  less  than  his  stock :  B's 
money  was  $50  more  than  A 's,  and  was  in  trade  5  months :  what 
was  A's  stock? 

Let         x  denote  the  number  of  dollars  in  A's  stock  ; 

x  +  50  "  "  «  B's     " 

x  —  60  "     A's  total  gain  ; 

x  —  60 

As  gain  per  month  ; 


3 

x  —  60 


"     A's     "  »    on  1  dollar ; 

50)   B's    "        «        « 


(— - — J  (x  +  50)  5  B's  total  gain. 

\  OiP  / 

From  the  conditions  of  the  problem, 

x  —  60  +  (*  ~  60)  (x  +  50)  5  =  140 ; 
clearing  of  fractions,  and  reducing, 

~~4~a 
By  the  rule, 


105625  _  325      475 

64      ~  ~8~      IT 


whence,  x  =  100. 


12.  Two  persons,  A  and  B,  start  from  two  different  points,  and 
travel  toward  each  other..  When  they  meet,  it  appears  that  A  has 
travelled  30  miles  more  than  B.  It  also  appears  that  it  will  take  A 
4  days  to  travel  the  road  that  B  had  come,  and  B  9  days  to  travel 


175.]  EQUATIONS  OF  THE  SECOND  DEGREE.  89 

the  road  which  A  had  come.     What  was  their  distance  apart  when 

they  set  out? 

• 
Let         x          denote  the  number  of  miles  B  travelled ; 

then  will     ar  +  30     "  "  "      A 

2 

A  travels  per  day ; 
x  + 30 


-—    «  «  days  A        " 

© 


(a: 

V 


-f  30 


«     B 


From  the  conditions, 


x         _  x_+  30  x2  _  x2  +  60*  -f-  900 

°r'    T~~      ~9" 


whence,  by  reduction, 

*2  —  48*  =  720 
and  by  the  rule, 


x  =  24  ±  -y/^20  +  576  =  24  ±  36  ; 
taking  the  upper  sign,        x  =  60,     and    x  +  30  =  90  ; 
hence,  the  distance  is  150  miles. 

EXAMPLES  INVOLVING  RADICALS  OF  THE  SECOND  DEGREE. 


CL  I  CL^1   '       (ifi  X 

3.  Given         -  +  \  /  -  -  —  =  T, 
x       V       xz  b 


to  find  the  values  of  x. 


90  KEY  TO  DAVIES'  BOURDON.  [175. 

Multiplying  botn  members  by  bz,  and  transposing, 

b  •y/a2  —  x2  =i  x2-  —  ab ; 
squaring  both  members, 

cancelling  62a2,  dividing  both  merr.bers  by  a2  and  transposing, 
x*  =  2ab-b*         .-.         x=  ± 


4.  Given       \  /_^_±_?  +  2 


Multiplying  both  members  by 

z  -f  a 

a; 

multiplying  both  members  by  a:,  and  transposing, 

2-y/aa;  =  b2x  —  x  —  a  =  (b2  —  l)x  —  a ; 
squaring  both  members, 

4ax  =  (b*  -  262  +  1)  xz  -  2a  (62  -  1)  x  +  a2 ; 
transposing  and  reducing, 


-,2 


rX  =    — 


5*  _  262  -j-  1  5*  —  262 

whence, 


/  a2  «2(6* 

1  d:  V  ~  6*  -  262  +  1  +   (6*  - 


-  262  -f  1    :        ~  6*  -  262  +  1        (6*  -  262  -f  1  )2  ' 


or, 


~  b*  —  2bz  +  1   ~  6*  -  262  +  1 ' 
now,  b*  -  2bz  -f  1  =  (b  -  1)2(6  +  I)2. 


175.]  EQUATIONS  OF  THE   SECOND   DEGREE. 

Hence,  taking  the  upper  sign  and  reducing, 


91 


_ 
= 


_q(64-l)2_         a(b  +  I)2  a 

=  ~  ~ 


(62  _  t)2  -  (b-  l)2(6  4-  I)2  ~  (b  -  I)2 ' 
and  taking  the  lower  sign  and  reducing, 

_  a(b  -  I)2  _         g(b  —  1)2  q 

~  (52  _  !)2  -  (b  _  1)2(6  4.  1)2  -  (j  +  i)a' 

or,  uniting  the  two  values  in  a  single  expression, 

a 

x  — 

(b  =F  i)2 

a  —  -J  q2  —  a;2 

5.  Given,         •  .  ;  =  0,     to  find  x. 

a  4-  -y/  a2  —  a;2 

Clearing  of  fractions,  transposing,  &c., 


squaring  both  members, 

02  (i  _  j)2  -  (6  4.  1)2  (a2  _  3.2) . 

transposing  and  reducing, 

2a  -^b 


(6  4-  I)2 


x—± 


6.  Given, 


—  a          7ia 


,     to  find  x. 


Multiplying  both  terms  of  the  first  member  by    */x  — 
and  then  dividing  both  members  by    a, 

I  n2 


2x  —  a  — 
clearing  of  fractions,  &c., 


a:  —  a 


—  a, 


92 


KEY  TO  DAVTES    BOUKDON. 


[175. 


(1  —  2n2)x  —  (1  —  n2)a  =  —  2w2  -y/a;2  —  ax  ; 
squaring  both  members,  transposing  and  reducing, 

x*  _  aiLrJ^a,  =  -.L=^±l*«». 

1     r   4-/i^  1          -    ^yj.2 

whence  by  the  rule, 

^   ~    1 ]J       9  "  \/  1  XI      9  '  7l 

1  —  4/i"2          V  1  —  4^  (1  - 


_ 

1 


2n3a 


-  4n2  4nz~  1  —  4n* 

Taking  the  upper  sign,  and  dividing  both  terms  of  the  fraction  by 

1  +2n, 
2n  +  na)a  _  (1  -  n)2q 


_ 


2n 


Taking  the  lower  sign,  and  dividing  both  terms  by  1  —  2n, 
+  2ra  +  ft2)  _  (1  +  n)2a 


_ 


+2n 


a  (I  ±  n}2 
taking  the  two  values  together,  x  =  —  ^  --  =-*•  •    • 


n-  V  ""    '     X    \     "Va  —  x  IX     t.      c.    j 

7.  Given          v — =  +     — =\/  T,  to  find  x. 

IX 


Multiplying  both  members  by  ^fx 


/a  +  x  -f-  i/o  —  a;  =  ~~=  > 
V* 


squaring  ooth  members, 


175.]  EQUATION*  OF  THE  SECOND  DEGREE.  93 

squaring  both  members, 


cancelling  4o2  in  the  two  members,  and  dividing  both  by  x*, 

_  x2      4a 
=  P~~b' 
clearing  of  fractions,  &c., 

z2  =  4ab  —  462  .  •  .  x  =  ±  2  Jab  —  &T 


a       x       -  ,          c    , 

8.  Given        ? —  £  •  to  find  x. 

a  -+•  x 

Clearing  of  fractions,  transposing  and  factoring, 


^/•2ax  +  x2  =  (b  —  1)  (a  +  x). 
Squaring  both  members, 

2ax  +  x2  =  (62  -26  +  1)  (a2  +  2a*  +  *2)  ;     or, 
2a*  +  z*  =  (62  —  26)  (a2  +  2a*  +  *2)  +  a2  +  2a«  -f- 
whence,  by  reduction, 

/I  _  26  + 


-— 
whence,  by  the  rule, 


«= 


- 
—  26 

a    1  - 


taking  the  upper  sign,       x  = 

< 

a(\ 
taking  the  lower  sign,      x  =  --  J 


whence,  a?  =  ± 


KEY  TO  DAVIES'  BOtTRDON.  [175-178. 

2d  Solution. 

Make  x  -f  a  =  y ;  whence,  2ax  -f  z2  =  y2  —  a* 

substituting  in  the  equation,  and  clearing  of  fractions, 

y  -f  yy  —  a2  =  iy  ; 
transposing,  &c., 

V>2  —  a2  =  (5  —  1)  y ; 
squaring  both  members,  and  cancelling, 

-  a2  =  (i2  -  26)  y2  ; 
whence,  solving  with  respect  to  y, 

a 

j     •"""     *•*-• 

substituting  for  y  its  value,  &c., 


x  =  —  a 


,    ,. 
whence,  as  before, 


^/26  -  &2 

r^±a(l-- 

-V/26  -  62 

TBINOMIAL   EQUATIONS 

6.  Given         x*  —  (2bc  +  4a2)  ar2  =  —  62c2  ;  to 
By  the  rule, 


4o*  =  ±e  +  2u2  ±  2av        f 
7.  Given        2«  -  7  y^  =  99  ;     or,    2z  -  99  =  7  ^/i 
Squaring  both  members 

4*2  —  396«  +  9801  =  49a? ; 


178-181]  EQUATIONS  OF  THE  SECOND  DEGREE.  95 

transposing  and  reducing, 


445  9801 

•*=-  — 


Ahcnce,  by  the  rule, 


445         /      9801   ,    198025 

ff   —  _L__f   __.    -^.  *     /    __    __  I          _  _  ,  _ 

*   ~~~       <"»        ~^   \  /  /*  j          J 


445       203 

or,  x  =  —  ^—: 

648 

taking  the  upper  sign,        x  =  -  —  =  81 

o 

242       121 

lower  sign,         x  =  — r-  =  —r— 

8   Given,  ^  —  bz*  +  C-x*  =  0,     to  find*; 

transposing  and  reducing. 


whence,  or  =  ±  1  ± 


reducing,  *  =  ±  J 

V 


2W 


EXAMPLES  OF  REDUCTION  OF  EXPRESSIONS  OF  THE  FORM  OF 


±  Vb. 


4.  Reduce  to  its  simplest  form,   \7^  ~ 

a  =  28,     5  =  300.     c  =  22. 


KEY  TO  DAVIES    BOUEDON.  [181. 

Applying  the  formula  and  considering  only  the  upper  sign, 


v/ 


28-f  10  1/3  = 


5.  Reduce  to  its  simplest  form,    */l  +  4^—3. 

o=l,         b  -  —  48,         c  =  7 ; 
applying  the  formula,  &c., 


+  4-v/-  3  =  2 
6.  Reduce  to  its  simplest  form, 


_  62  _  ^Jbc  _ 


a  =  ic,         i  =  462  (6c  -  62),         c  =  b  (c  -  2b)  ; 
applying  the  formula  to  the  1st  radical, 


+  26  -vA  -  *8  = 
applying  the  formula  to  the  2d  radical 


subtracting  the  second  result  from  the  first, 


6  ytc  —  b2  —bc  —  2b  -foe  —  b2  =  ±  2A. 
7    Reduce  to  its  simplest  form, 


\/ab  -r  4c2  —  d2  —  2 
a  =  aft  -f  4c2  —  <P,       6  =  16a5c2  -  4a£flP,       c  =  ab  —  4C2  f-  «PV. 


181-183.]     EQUATIONS    OF    THE    SECOND    DEGREE.  97 

Applying  the  formula, 


—  ob®  =  ± 

SIMULTANEOUS  EQUATIONS. 

(x  +  2y  =  7  .........     (1) 

^  =  31    ......      (2) 


From  (1), 

x  =  7  —  2y. 
Substituting  in  (2), 


49  -  2%  +  V  +  3y  (7  -  2y)  +  #2  =  31  ; 
hence, 


_7  ,4/49      18 _    _7      11 
2*  '    4     ~  4  ~        2±2: 

.-.     y'=Z        and        y"=— 9; 


and 

x'=  3        and        «"=  25. 

(2z  +  y  =  27 (1) 

(3a:y  =  210 (2) 

From  (1), 

y  =  M  —  2x. 

Substituting  in  (2), 

3z(27  —  2z)=210; 
hence, 


27 

=  T±y^;-3o  = 

.-.    x'=W        and        x"=  3f 
From  (1) 

y=  7        and        y"=  20. 

4  (2x  —  3y  —  1  =  0 (1) 

1  /  r^  \ 

From  (1), 

*  =  — 2~  J 


98  KEY    TO    DAVIES'    BOURDON.  [183-184. 

from  (2), 


hence, 


_  7.L.449   .   312  _  -7  ±19 
±T        +         " 


4  16  16  "          4 

.-.     y'=3        and  y"=-6J; 
and  from  (5), 

«'=  5  and  «"=  —  9£. 

{Wx  +  y 
-^  =  3   ......  -.     (1) 
xy 
y-x  =  2  ........    (2) 

From  (2), 

y  =  2  +  x. 

Clearing  and  substituting  in  (1), 


hence, 


SG"       6 

.-.      x'=2        and        «"=—  |. 
From  (2), 

y'=4        and        y»=| 


(1) 

•    •      (2) 

Adding  (1)  and  (2)  and  dividing  by  2, 

o«=  36;        .-.    x'=6       and       x"=  —  6; 
y'=  5        and        y"=  —  5. 

(1) 


184.]  EQUATIONS    OP    THE    SECOND    DEGREE.  99 

Making  y  =px, 

4.8 

*-p*  =  48;    .-.    #  =  ....     (3) 


Equating  (3)  and  (4)  and  dividing  by  12, 
4  1 

1  — p      p — p*} 
hence, 

51  5   ,    ,/25 16 

?-lP  =  -l'>    •'•    P  =  Z±VU-& 

or,  p  =  %    and    p  =  ~. 

4 

Using,    p  =  -,    we  have  from  (3), 


x'=  +  8    and    x"=  —  8;    whence,   y'=  -f  2    and    y"=  —2. 

9. 
Making 


9  j  s2  +  4^  +  4^  =  256 (1) 

(3yz  —  a^  =  39 (2) 


x1  -\-  kpx*  +  4j02.r2  =  256 (3) 

3p*x*  —  x*  =  39 (4) 

From  (3)  and  (4), 


1  -(-  4p 
Equating, 


256  39 


ufi&06 

768P2  -  256  =  39  +  156jo  +  156J03, 

13         295 
and  P2  — ^rP=7^r; 


KEY    TO    DAVTES'    BOURDON.  [184 

hence, 

78  ±  432  5  59 

p--    -eia-'  or  ^  =  6   and  *  =  -ioa' 

Using  the  first  value,  we  have  from  (6), 

a;2  =  36;         •.        3'=  6        and        x"=—  6; 
and  by  substitution, 

y'=  5        and        y"=  —  5. 
Using  the  second  value,  we  find, 

a'  =102        and        x"  =—102; 
and  by  substitution, 

y'=  —  59        and        y"=.  59. 


10  J6(a?  +  ^)  =  13ajy  ......     (1) 

(z2  —  2/2  =  20  ........     (2) 

Making 


6(1  +^2)a;2=:  13^2  ......     (3) 

(1-^2)^  =  20     .......     (4) 

From  (3),  we  have, 

6  +  6p2  =  I3p. 

Reducing, 


_  13  .  /169  _  144  _  13  ±  5 

'*•    ^  ~  12  ^  144  ~~  144  ~       12     ' 
or 

^  =  H  and       p  =  -. 


188.]  EQUATIONS    OF    THE    SECOND    DEGREE.  101 

Using  the  second  value  of  p,  we  have  from  (5), 

on 
a?  =  T--=4  =  36,     or    x'=  6    and    x"=  —  6 ; 

1  —  f 
and  by  substitution,    y'—  4    and    y"=  —4. 

EQUATIONS  OF  A  HIGHER  DEGREE  THAN  THE  FIRST,  INVOLVING 
MORE  THAN  ONE  UNKNOWN  QUANTITY. 

(  x2u   +  xv2   =    6  ...  (1)  ) 
15.  Given,          •<  f-    to  find  x  and  y. 

tzy +  zy  =  i2.. .  (2)> 

Dividing  (2)  by  (1),  member  by  member, 

2 

xy  =  2,     or     a;  =  — 

Substituting  this  value  of  x  in  (1)  and  reducing, 

*  +  *  =  •; 

clearing  of  fractions  and  reducing, 

y2  -  By  =  -  2 : 


whence,  y  =  |  ±1/-  2  +  i  =  i  ±  I; 

or,  y  =  2,     and     y  =  1 ; 

whence  a?  =  1,     and     a;  =  2. 

16.  Given,      J  *'  +  *  +  y  =  18  ~  y2  '  '  '  0)  It  o  find  ,  and  y. 
(  xy  —  6 (2)  ) 

Multiplying  both  members  of  (2)  by  2  and  adding  the  resulting 
equation  to  (1),  member  to  member, 

z'  -f  2*y  +  y2  +  *  \-y  =  30, 
or,  (x   +  y)2  +  x  +  y  =  30  ; 


102  KEY  TO  DAVIES'  BOURDON.  [189. 

whence,  by  the  rule, 


whence,  x  -\-  y  =  5,     and    x  +  y  =  —  6. 

Taking  the  first  value  of  x  +  y  and  substituting  in  it,  for  y  its 

/» 

value  —  derived  from  (2), 
x 

x  +  —  =  5; 

x 

clearing  of  fractions  and  reducing, 

xz  —  5x  =  —  6 ; 


.  5          /      _   ,    25       5  ^  1 

whence,  a.  =  _±y/_6  +  —  =  -±-; 

ur,         x  =  3,     and    #  =  2  ;     whence,    y  =  2,     and    y  =  3. 
Taking  the  second  value  of  x  +  y  and  proceeding  as  before, 


clearing  of  fractions,  &c., 

a;2  +  6z  =  —  6  ; 


whence,  x  —  —  3  ±  /—  6  +  9  =  —  3 

and  by  substitution,  y  =  —  3 


PROBLEMS  GIVING  RISE  TO  EQUATIONS  OF  A  HIGHER  DEGREE 
THAN  THE  FIRST  CONTAINING  MORE  THAN  ONE  UNKNOWN 
QUANTITY. 

2.  To  find  four  numbers,  such  that  the  sum  of  the  first  and  fourth 
shall  be  equal  to  2s.,  the  sum  of  the  second  and  third  equal  to  2/, 
the  sum  of  their  squares  equal  to  4c2,  and  the  product  of  the  first 
%nd  fourth  equal  to  the  product  of  the  second  and  third. 


189-191.]        EQUATIONS   OF  THE   SECOND   DEGREE.  103 

Assuming  the  equations, 

«  +  2  =  2*  •     •     •     •  (1) 

x+y  =  W  ....  (2) 

w2  -(-  z2  +  y2  +  z2  =  4c2  .     •     .     •  (3) 

uz  =  xy  •     •     «     •  (4) 

Multiplying  both  members  of  (4)  by  2,  and  subtracting  from  (3), 
member  from  member, 

u2  —  %uz  -f-  z2  +  a;2  -f  2ary  +  y2  =  4c2 ; 
or,  («  —  z)2  +  (x  4-  y)2  =  4c2. 

Substituting  for  x  +  y  its  value  2s'  and  transposing, 

(u  -  z)2  =  4c2  -  4s'2, 

or,  u—  z—  -v/4c2  —  4s'2. 

Combining  with  (1), 


,/4c2  -  4*'2  

and,  z  —  s  —  v- ~    -  —  s  —  ^&  —  s'2 ; 

reversing  the  order  of  the  members  of  (4),  and  proceeding  as  before, 
we  find  in  like  manner. 


and, 

4.  The  sum  of  the  squares  of  two  numbers  is  expressed  by  a, 
and  the  difference  of  their  squares  by  b  :  what  are  the  numbers? 

Let  x  and  y  denote  the  numbers. 
From  the  conditions  of  the  problem, 


J04:  KEY  TO  DAVIEs'  BOURDON.  [191. 

x2  +  y2  =  a     •     -     •     •     (1) 
z*-y*  =  b     •     •          .     (2). 
By  adding,  member  to  member, 


by  subtracting, 


5.  What  three  numbers  are  they,  which,  multiplied  two  and  two. 
and  each  product  divided  by  the  third  number,  give  the  quotients 
«,  6,  c  ? 

Let  #,  y  and  2,  denote  the  numbers 
From  the  conditions  of  the  problem, 

-  =  a     or,     xy  =z  02     «     •     .     (1) 

P 

—  =  b     or,     yz  =  bx     •     •     •     (2) 

#2 

—  =  c     or,     xz  =  cy     •     •     •     (3). 

Multiplying  (1),  (2)  and  (3)  together,  member  by  member, 

xzyzz2  =  abcxyz  ; 
dividing  both  memoers  by  xyz, 

xyz  =  abc     •     •     •     (4)  ; 

substituting  in  (4)  the  value  of  xy  taken  from  (1).  and  dividing  both 
members  by  a, 

2?  =  be  .  •  .  z  = 

Substituting  the  value  of  yz  and  dividing  by  b, 


191.]  EQUATIONS  OF  THE  SECOND  DEGREE.  105 

x2  =  ac  .  '  .  x  =  Joe. 

Substituting  the  value  of  xz  and  dividing  both  members  by  c, 
•yz  =  ab  .'.  y  — 


6.  The  sum  of  two  numbers  is  8,  and  the  sum  of  their  cubes  is 
152  ;  what  are  the  numbers  ? 

Let  x  and  y  denote  the  numbers. 

t 

From  the  conditions, 

x  +  y  =      8     .....     (1) 
z3  +  y3  =  i52     .....     (2); 
cubing  both  members  of  (1), 

x3  +  3z2y  +  3xy*  +  y3  =  512     •     •     •     •     (3)  ; 

subtracting  (2)  from  (3),  member  from  member,  and  dividing  both 
members  by   3, 

xzy  +  xyz  =  120     •     •     •     •     (4)  ; 

- 

substituting  the  value  of  a;  taken  from  (1), 

(64  -  16y  +  y2)  y  +  (8  -  y)  yz  =  120, 
or,  reducing,  y2  —  Sy  —  —  15  ; 


whence,      y  =  4  ±  -y/—  15  +  16  =  4  ±  1     .  • .     y  =  5,     y  =  3; 
whence,  from  (1)  a;  =  3,     a?  =  5. 

7.  Find  two  numbers,  whose  difference  added  to  the  difference  ol 
their  squares  is  150,  and  whose  sum  added  to  the  3um  of  theii 
squares,  is  330. 

Let  x  and  y  denote  the  numbers. 
From  the  conditions  of  the  problem, 


106  KEY  TO  DAVIES'  BOURDON.  [191. 

a?_y2  +  *-y=150      •      •      •      •      (1), 

'   y     *^       »   y  \  J  t 

adding  member  to  member,  and  reducing, 

x2  +  x  =  240 ; 
whence,  x  =  —  -  ±  -/240  -f  ^  =  —  ^  ±  y  J 

or,  considering  only  the  positive  solution,  . 

x  =  15; 
whence,  from  (1),  by  substitution, 

8.  There  are  two  numbers  whose  difference  is  15,  and  half  their 
product  is  equal  to  the  cube  of  the  lesser  number :  what  are  the 
numbers  1 

Let  x  and  y  denote  the  numbers. 
From  the  conditions  of  the  problem, 

*-.y=15      •     •     •     (1), 


f    •  -    •    (2); 

qp 

substituting  in  (1)  and  dividing  both  members  by  2, 

2_y  _  15. 
y       2~  2  ' 
whence, 

15      J_  _  1       11 

considering  only  the  positive  solution, 

y  •=  3  ;     whence,  from  (1),     x  =  18. 


191-192.]  EQUATIONS  OF  THE  SECOND  DEGBEE.  107 

9.  What  two  numbers  are  those  whose  sum  multiplied  by  the 
greater,  is  equal  to  77 ;  and  whose  difference,  multiplied  by  the 
lesser,  is  equal  to  12? 

Let  x  and  y  denote  the  numbers. 
From  the  conditions, 

(x  +  y)x  =  n,     or     z*+'zy^W     .     .     .     (1)  ; 
(x-y)y=l2,     or     xy  -  y*  =  12     .     .     .     (2); 
make     x  =  py  ;     whence, 

Ttt     or,     ya  =  -JL-    .     .     .     (3), 


(P   -I)y2=12,     or,     y*  =  ——j    .     .     .     (4); 

equating  the  second  members  and  reducing, 

2_65  _77i 

P        l'2P  ~       12' 


65          /     77   ,   4225       65       23 
whence,        p  = —  ±^/ - -+ —  =  —  ±—. 

taking  the  upper  sign, 


/Qi*  O 

substituting  k  (4), 


whence,  x  =  —  -y/2~; 


i.       i.    ,  42      21 

taking  the  lower  sign,          p  —  —  =  —  ; 

<w4  1  ^ 


108  KEY  TO  DAVIES*  BOURDON.  [192. 

substituting  in  (4), 

/T44 
y  =  \J  —Q-  —  4  ;      whence,      x  —  7. 

10.  Divide  100  into  two  such  parts,  that  the  sum  of  their  square 
roots  may  be  14. 

Let  x  and  y  denote  the  parts. 
From  the  conditions, 

x  +      y  =  100    ...     (1), 

y7+  VF  =    14     ...     (2): 

squaring  both  members  of  (2)  and  subtracting  (1),  member  from 
member. 

2y^=96,     or     */xy  =  48,     or    xy  =  2304 : 
substituting  for  y  its  value,  100  —  x, 

100*  -  z2  =  2304,     or    x*  —  WQx  =  -  2304. 
whence,  by  the  rule, 

x  =  50±yr196"  =50  ±14; 
hence,  x  =  64,     x  =  36, 

and  y  =  36,     y  =  64. 

11.  It  is  required  to  divide  the  number  24  into  two  such  parts, 
that  their  product  may  be  equal  to  35  times  their  difference. 

Let    x  and  y     denote  the  parts. 
From  the  conditions  of  the  problem. 

*  +  y=:24     .     .     .     (1) 

ay  =  85(*-y)    .     .     .    (2); 

substituting  in  (2)  the  value,     y  =  24  —  a-, 

24*  -  ar2  =  35  (2*  —  24),     or,     24*  -  .c2  =  70 x  —  840 ; 


192.]  EQUATIONS  OF  THE  SECOND  DEGREE.  109 

whence,  z*  +  46z  =  840  ; 

by  the  rule, 

*  =  —  23  ±  ^840  +  529  =  —  23  ±  37  ; 

ence,  taking  the  upper  sign,     x  =  14  ; 
oy  substitution,     in  (1),  y  =  10. 

12.  What  two  numbers  are  those,  whose  product  is  255,  and  the 
sum  of  whose  square  is  514? 

Let  x  and  y  denote  the  numbers. 
From  the  conditions,         xy  =  255     •     •     .      (1) 
xz  +  y>  =  514  .     .     (2), 

multiplying  both  members -of  (1)  by  2,  adding  and  subtracting  the 
resulting  equation  to  and  from  (2),  member  by  member, 

ac*  -f  2xy  +  7,2  =  1024     .     .     (3) 
*2-2zy  +  y2  =  4      .     .     .     (4); 
extracting  the  square  root  of  both  members, 
x  +  y  =  32, 
*-y  =  2, 
whence,  x  =  17 ;        y  =  15. 

13.  There  is  a  number  expressed   by  two  digits,  which,   when 
divided  by  the  sum  of  the  digits,  gives  a  quotient  greater  by  2  than 
the  first  digit ;  but  if  the  digits  be  inverted,  and  the  resulting  num- 
ber be  divided  by  a  number  greater  by  1  than  the  sum  of  the 
digits,  the  quotient  will  exceed  the  former  quotient  by  2 :  what  is 
the  number? 


110  KEY  TO  DA  VIES'  BOURDON.  [192. 

Let  x  and  y  denote  the  digits  ;  then  will 

lOx  +  y  lenote  the  number. 
From  the  conditions, 


clearing  of  fractions,  and  reducing, 

Sx  —  y    -  xz  -  xy  =  0     •     •     «    (3), 
6y  —  4z  —  xz  —  xy  =  4    •     •     •     (4) ; 

by  subtraction, 

12*-f4 
•fy-12*  =  4;        .-.     y  = ; 

substituting  in  (3), 

12*  +  4  12^  +  4«_ 

0>c  «  7  ~ 

clearing  of  fractions  and  reducing, 

40  4 

-*=- 


_20         /      4_      400  _  20      18 
"19      V       19  +  361  ~  19       19* 

Taking  the  upper  sign,  gives    x  =  2 ;     whence,     y  =  4. 

14.  A  regiment,  in  garrison,  consisting  of  a  certain  number  of 
companies,  receives  orders  to  send  216  men  on  duty,  each  company 
to  furnish  an  equal  number.  Before  the  order  was  executed,  three 
of  the  companies  were  sent  on  another  service,  and  it  was  then 
found  that  each  company  that  remained  would  have  to  send  12  men 
additional,  in  order  to  make  up  the  Complement,  216.  How-mam 


192.]  EQUATIONS  OF  THE  SECOND   DEGREE.  Ill 

companies  were  in  the  regiment,  and  what  number  of  men  did  each 
of  the  remaining  companies  send  ? 

Let  x  denote  the  number  of  companies,  and 

y      "       "         "  each  should  send ;  then, 

y  +  12     will  denote  the  number  sent  by  each. 

From  the  conditions  of  the  problem, 

*y  =  216     •     •     •     (1), 
(*-3)  (y+12)=216     •     •     .     (2); 
Performing  operations,  subtracting  and  reducing, 

4x  —  y  =  12.  .  • .      y  =  4a;  —  12 ; 

substituting  in  (1),      4z2  -  12*  =  216,     or    **  —  3z  =  54 ; 

whence, 

taking  the  upper  sign, 

x  =  9  ;        hence,        y  =  24,     and    y  +  12  =  36. 

15.  Find  three  numbers  such,  that  their  sum  shall  be  14,  the 
sum  of  their  squares  equal  to  84,  and  the  product  of  the  first  and 
third  equal  to  the  square  of  the  second. 

Let  x,  y  and  z  denote  the  numbers. 
From  the  conditions  of  the  problem, 

x  +  y  +  z   =  14    •     •     .     (1), 
*2  +  y2  +  *2  =  84     .     •     •     (2), 
yz  =  y2     •    •     •     (3). 
Multiplying  both  members  of  (3),  by  2,  adding  to  (2)  and  -educing, 


KK1  TO  DAVIES*  BOURDON.  [192. 

from(l),  x  +  z  =  U-y  •  •  •  •  (5); 

equating  the  second  members  of  (4)  and  (5)  and  squaring, 

84  +  y2  -  196  —  28y  +  y2 ;         .  • .     y  =  4. 
Substituting  in  (1)  and  (3), 

x  +  z  =  10     .     .     .     .     (6) 

16 
xz  =  16    .  •. 


Substituting  in  (6)  and  reducing, 

22  _  10*  =  -  16 

2  =  5  ±  ,./9~=  5  =fc  3, 

V  *M 

0  =  8;         a  =  2, 
and  by  substitution,        x  =  2  ;         a;  =  8. 

16.  It  is  required  to  find  a  number,  expressed  by  three  digits, 
such,  that  the  sum  of  the  squares  of  the  digits  shall  be  104;  the- 
square  of  the  middle  digit  to  exceed  twice  the  product  of  the  other 
two  by  4  ;  and  if  594  be  subtracted  from  the  number,  the  remainder 
will  be  expressed  by  the  same  figures,  but  with  the  extreme  digits 
reversed. 

Let  a?,  y  and  z  denote  the  digits ; 
then,  lOOx  +  lOy  +  *    will  denote  the  number. 

From  the  conditions  of  the  problem, 

z2  +  y2  +  *2  =  104     .     .     .     (1) 
yz  —  Zxz  =  4         .     .     .     (2) 
100*  +  10y  +  z  -  594  =1002  +  lOy  4-  x    (3) ; 


193.]  EQUATIONS  OF  THE  SECOND  DEGREE.  113 

subtracting  (2)  from  (1),  member  from  member, 

x2  -i  2xx  +  z2  -  100        .  •  .        x  +  z  —  10  ; 
reducing  (3)  x  —  z  =  6  ; 

hence,  x  =  8,     and    z  =  2. 

By  substitution,        y  =  6,     and  the  number  is  862. 

17.  A  person  has  three  kinds  of  goods  which  together  cost 
$230^.  A  pound  of  each  article  costs  as  many  fa  dollars  as  there 
are  pounds  in  that  article  :  he  has  one-third  more  of  the  second  than 
of  the  first,  and  3^  times  as  much  of  the  third  as  of  the  second  . 
How  many  pounds  has  he  of  each  article  ? 

Let  ar,  y  and  z  denote  the  number  of  pounds  of  each  article. 
From  the  conditions  of  the  problem, 

or   x"  +  *  +  *  =  6525   1 


4  16 

y  =  \*  .'.  2/2  =  y2    .-.    (2) 

2  =  -y       .-.     z=-2%,      and,     z2  =  —  z2      ...     (3)  ; 

substituting  in  (1)  and  reducing, 

*2  =  225          .'.    .        x  =  15, 
substituting  in  (2)  and  (3), 

y  =  20  ;  z  =  70. 

18.  Two  merchants  each  sold  the  same  kind  of  stuff:  the  second 
sold  3  yards  more  of  it  than  the  first,  and  together,  they  received 
35  dollars.  The  first  said  to  the  second,  "  I  would  have  received  24 
dollars  for  your  stuff."  The  other  replied,  "And  I  would  have 


114  KKY  TO  DAVTES'  BOURDON.  [193. 

received    12^  dollars  for  yours."     How  many  yards  did  each  of 
them  sell? 

Let  x  and  y  denote  the  number  of  yards  sold  by  each. 

24 

Then  will  —  denote  the  price  the  first  received  per  yard, 

25 
and  Q-  will  denote  the  price  the  second  received  per  yard. 

From  the  conditions  of  the  problem, 
*  +  3  =  y, 


+       .  =  35,     or,     4S*2  +  25y2  =  ?0zy  : 

substituting  in  the  second  equation  the  value  of  y  taken  from  the 

first, 

48*2  +  25  (xz  +  Qx  +  9)  =  70*2  +  210*  ; 

reducing,  x2-  —  20x  =  —  75  ; 

whence,  x   =  10    ±  -y/25^  =  10  ±  5, 

/ 

or,  ar   =  15  ;  x  =  5  ; 

substituting,  y  =  18  ;  y  =  8. 

19.  A  widow  possessed  13000  dollars,  which  she  divided  into 
two  parts,  and  placed  them  at  interest,  in  such  a  manner,  that  the 
incomes  from  them  were  equal.  If  she  had  put  out  the  first  portion 
at  the  same  rate  as  the  second,  she  would  have  drawn  for  this  part 
360  dollars  interest  ;  and  if  she  had  placed  the  second  out  at  the 
same  rate  as  the  first,  she  would  have  drawn  for  it  490  d(  liars 
interest.  What  were  the  two  rates  of  interest  ? 

Let  x  and  y  denote  the  rates  per  cent. 

Let  z  denote  the  1st  portion  ;  then  will  13000  —  z  denote  the  2d. 


224.]  EXTRACTION   OF   ROOTS.  115 

P  roni  the  conditions  of  the  problem, 
xz        (13000  —  z\  y 


or,        »  =  1300<*-*     •         •     (1), 
j^  =  360,  or,         2y  =  36000  .     .     .     (2), 

p^-?  =  490,        or,         13000*  -zx  =  49000  •     •     (3). 

Substituting  in  (1)  the  values  of  zy  and  zx  taken  from  (2)  and  (3) 
and  reducing,  we  find,  x  =  y  +  1. 

Substituting  this  value  of  x  and  the  value  of  z  taken  from  (2)    in 
(1),  and  reducing,  we  find 

72         36 
v   --  v  —  —  : 
y        13  y       13' 


36          /36       1296       36  ±  42 
whence,       y    -.=  -  ± ^ -  +  —  =     __;         .-.     y  =  6 ; 

by  substitution, 

*  =  7,     and    «  =  6000,     13000  -  2  =  7000. 

ADDITION  AND  SUBTRACTION  OP  RADICALS. 

2.  3  A/4a2 : 

3.  2  A/45  =  6  A/5;    3A/5,  .'.   ^4/w.  9A/57 


4.  3a  V*  —  2c  A/*,  .'.   Ans.  (3a  —  2c)  V*- 

5.  3  Vi«2  —  2  V2a ; 

3A/4^  =  3A/2a;  .'.   Ans.    V% 

6.  A/243  =  9  A/3;  A/27  =  3\/3;  A/48  =  4  A/3.    -4ws.  16  A/3. 


116 


KEY  TO  DAVIES'  BOURDON.  [224-225. 


7. 


5  A/7205  = 


=  —  56 

.  (13a  -  56)  VSai 


8. 


9. 


10. 


5. 


6. 


V64 


16a4  =  2a  yb  +  2a,    and 
+  2«, 

«;    .'.     Ans. 


MULTIPLICATION  OF  RADICALS. 

X      3  =  1/64  x 


*  /r   3  /r  i2  /r   «  /T    12  /" 

V  2  X  V  3  =:   V  8  X    Vil=!   V  648  ;     '  '•  ^  " 


6   /2         10 

-  VI-4 


^;  3      10"=  3/100;  .-.  ^ns.  6      337500 
10 


8. 


.         048000. 


7/4      42  /4096     3/1      42/i 

O      ./       t   / .      4/_  —     «/ .      14  /ft  

'VI        V  729  '    V2~   V  10384'     V° 

. ' .     Ans. 
10.  The  product  by  the  rule  for  multiplication  is 

28   .          ft   .          n   .    10      43   .  H      43  .    13 


42  /2 


226-227.]          TRANSFORMATION    OF    RADICALS. 

9w2  3m 


117 


11. 


12.  The  product  equals 


3m  "  3m 

.-.    Ans.  =  Va?  +  P. 


DIVISION  OF  RADICALS. 

2.  2  V3  x  V4  =  2  v^729  x  1<v/256  =  2  ^729  x  256, 
iA/2  X  V3  =  i1'V/8  X  1'V/81  =  i1v/8^^1 

2  J^729  x  256  --  i  ^8^81  =  4  V  =4^288. 


j  _          _ 

4.     —  =  --  =  :  multiplying  both  terms  by  i/a  —  */b,  we  hav« 

Va  +  V* 


again  multiplying  both  terms  by  -y/a  +  V^  we  finally  obtain 

\f  \J  I       ^/  •*/  j 

—        Ans. 


a  —  b 


multiplying  both  terms  by  */a  -f  */6,  we  have 

multiplying  both  by   -y/a  4-  V^T  an^  we  have 
o  +  6  4-  2  i/o2 


a-6 


118  KEY   TO    DAVIES'    BOUEDON.  [227-239. 

6.  V43  =      86  + 


Multiplying  both  terms  of  (1)  by  3  A/3T  —  6  A/3, 

173  A/3T  +  26  A/31  x  43  —  516  V3  —  78  A/84 
84-73 

26V31  X  43  =  546  A/3;     —  78  A/84  =—  156  A/31  ; 


13 

FRACTIONAL  EXPONENTS. 


1.  alTc~l  multiplied  by  a? 

4  +  a=l-5     -2  +  3 
hence,  a$lTzc~l  x  a2fr 

3.    3cr2^    multiplied  by    % 

4  14       3      1      7 

n  l^  .  _i  •         ft_l_O  _  9. 

~~     "'         +     ~' 


hence,  3a~25 

3.    6a~^Z>4c~m     multiplied  by 
--  +  -=—-;    4  —  5  =  —  1;     —  m  +  w  =  w  —  m; 


hence,  6a~4c-m  x 

4.    (q«^)    multiplied  by    ^a*. 

3341  3 

—  v  —  —  —  *       —  v  2  —  —  • 

3X3~9'      3X      ~3' 
hence, 


239.]  TKANSFOBMATION    OF    RADICALS. 

1       1       1_  JL.      1       1       !_?. 

3  X  3  X  3  ~  27 '      o  +  o  +  o  —  o  5 

(q6 

6.    at    divided  by    a~*. 


\3      ! 
hence,  -<,*=  _« 


j      3_17. 
S    i  V    4l~~3      4~12' 

0  ,  11 

hence,  at  ^-  at  =  a1^. 

7.  a*    divided  by    «». 

3_4_         1 
4      5~        120J 

hence,  a*  -j-  a^  =  a~^V. 

8.  a»  x  i*    divided  by    a~ zb» 

2_/l\_  2      1__9_.      3_7_    _1. 
5~\2/~~  5  +  2~10'      4      8~        8' 

hence,  a*b*  -$-  a~%b°  =  flifrJ  °. 

9.  32a^J6c^    divided  by 


=    ;       - 

hence,  32aii0 

10.  64a9jVt    divided  by 
64-32  =  2;  9-(-9)=18;  |- 

hence,  64a9jV^  -r-  32a-»5~  *c~*  =  4a18i5. 

11.  VS  x  Vji  x 


hence,  x  x 


12.  Reduce    ?     ,  x  to  its  simpiest  terms. 


120  KEY  TO  DA  VIES'  BOUEDON.  [239. 

Cancelling  the  <y/2~,  and  writing  the  quotient  of  2  -f-  J,    we  have 

43^37    Ans. 

C  i      -  \  * 

I  4  (2\*  3/3 

13.  Reduce       1  —  —    "*       V      to  its  simplest  terms. 


liaising  both  terms  to  the  4th  power,  we  have 

(i)*g"8*  _  (i)M3)V3        /V3  J_, 

(2)*  2  (3)8  ~  (S)'3  (3)2  (2)7  x  3  ~  384     0' 


U.  Reduce     /J  fl2±J$  t? 


to  its  simplest  terms. 


Since  the  square  root  of  the  square  root  is  equal  to  the  4th  root, 
we  need  only  operate  on  the  terms  of  the  fraction  : 


Multiplying  both  terms  of  the  fraction  by  the  -\/6^  we  have 


:  hencej 


15.  Multiply     a^+  a2^  +  A   +  aJ  +  oV  +  6* 


8  414 

g*b  -  ab*  -  aV 


3  — 


240.]  TRANSFORMATION    OF    RADICALS. 

i  8  -1-  1 

16.    Divide    a?  —  azb~*  —  cfib  +  5 


17.    a;*-£a* 


fa 


&c.,        &c. 


18. 


y    +  y?y*      +  x 


+  aTjj    +  1 


&c. 


121 


10  _         9 

~3  /~4  "     ~6~    "  9 


20. 


+ 


+  y~l 


ar«  +  «V7  + 

_3    _1  _       _ 

+  af  V*  + 


+  z- 


122 


KEY  TO  DA  VIES'    BOUKDON. 


21. 


22.    x   —  ±x    +  2 

—* 


X     —X 


2x 


Qx  —  x*  —  2x. 


23.  ix'tt— 


[240. 


*—* 


24.    x*+  xy  +  y2 
a;  +  a?*y*  + 


+  dly  + 


25.    a^  +  2a£  +  3^ 
g^  —  2x^  -f-  1 


240-248-249.]  ARITHMETICAL  PROGKESSION. 
26.  a?  +  a;*  —  1|&  —  f  a;*  +  $    1  x  +  ja£  - 


123 


+  f 


ARITHMETICAL  PROGRESSION. 

2.  a  =  2,     d  =  7,     7i  =  100;     .  •.   f  =  a  +  (n  -  l)rf  =695. 

3.  a  -  1,     rf  =  2,     w  =  100  ;     .  •  .    I  =  a  +  (n  -  1  )  d  =  199. 
Hence,  S  =  $n  (a  +  I)  =  10000. 

4.  I  =  70,     rf  =  3,     »  =  21, 

a  =  ?  _  (w  _  1)  d  =  10  ;     S  =  ln[2l-(n-  l)d]  =  840. 

5.  a  =  10,     d=—  J,     w  =  21     .  •.  Z  =  a  +  (n  —  l)d  =  J£'t 
whence,  S  =  %n  [21  —  (n  —  l)d]  =  140. 

6.  d  =  6,     1=  185,     5  =  2945 


_ 


21  +  d  ±  -y/(2/  H-  rf)2  -~8dS_  376  ±4 


taking  the  lower  sign,     ?i  =  31,         a  =  /  —  (^^  —  \)d  =  5. 


7.  a  =  2,     f  =  5,     n  =  ll;     .  •  .  d  = 

8.  o=l,     rf  =  1,     n  =  n     ^  =  n; 


n  —  1       10 


=       =  0.3. 


SB 
9.     a  =  1,     c?  =  2,     n  =  n  ;     .  •  .   5  =  ^n  [2a  +  (n  —  l)rf]  =  wj 

10.     o  =  4,     o?  =  4,     n  =  100  ; 

.    .     S  =  %n  [2a  +  (»  -  1)^]  =  20200 


124  KEY  TO  DAVIES'  BOUKDON.     [253-54-56-58. 

GEOMETRICAL   PROGRESSION. 
3.     a  =  2,     r  =  3,     I  =  39366, 

=  59048. 


r  —  1 

4.  a  =  l,     r  =  2,     n  =  12; 

ft  fit     __    /Y 

...     £  _  S.  -  J  =  4095  ;     Z  =  or—  i  =  2048. 
r  —  1 

5.  a  =  1,     r  =  2,     n  =  12  ; 

qr*  ~ 


.. 

r  —  1 

6.     o=l,     r  =  3,     n  =  10  ; 


=  4095  ;  4  )95.?.  =  £204  15*. 


r-l          2 

I  =  or"-1  =    196.83. 

INSERTING   GEOMETRICAL   MEANS. 

2.     a  =  2,    J  : 

hence,  the  progression,        2  :  6  :  18  :  54  :  162  :  486. 

SUMMATION   OF  SERIES. 
4.  p  =  4,     q  =  4,     n  =  1,     5,     9,     13,     &c. 

4,4,4,4,4, 
1st  auxiliary  series,     7  +  ^  +  0  +  70  +  77+  *  *  * 

4444  4 

2d  +  e   +  K  +  To  +  Trv  +    *  *  *  + 


5   '   9   '    13   '    17   '  '   4w-3 


287-311.]  EXPONENTTAX,    EQUATIONS.  125 

PILING  BALLS. 


2.    .. 


w'  =  5,     £'  =  55  ;       .-.  S—  S'  =  960. 


S.  m  =  30,    .  =  »!      .^=!1<?1±1}  (1+^=23405. 

I  •  iff  .8 

4.  m  =  26,     n  =  20  ;       .  •  .  5  =  8330  ;     m  =  26,     n'  =  8  ; 

/&"  =  1140;      .'.5-  £'  ^7190. 

5.  w  =  20  ;     .  •  .  S  =  1540  ;     n'  =  9  ;     .  '.  S'  =  65  ; 

.  •  .  S  -  S'  =  1475. 

6.  n  =  15  ;     .  •  .  S  -  1240  ;     n'  =  5  ;     .  •  .   S'  =  55  ; 

.-.  S-  5'  =  1185. 

7.  m=  52,     n  =  40  ;     .-.  S=  C4780  ;     m  =  52,     nx  =  18  ; 

.-.£'  =  11001;     .-.  S  -  S'  =  53679. 

EXPONENTIAL  EQUATIONS. 

These  equations  may  be  solved  as  the  preceding  ones  have  been, 
but  it  will  be  better  to  make  use  of  the  table  of  logarithms  (P.  291), 
8*  =  32  ;    hence  we  have, 

log  32      5 
;    or,    x  =  ^  =  ^ 

2.  3*  =  15  ; 

taking  the  logarithms  of  both  members, 

log  15       1  .  176091 
. 


1 26  KEY  TO  DAVIES'  BOURDON.  [311-331. 

3.  10*  =  3; 

taking  the  logarithms  of  both  members, 

x  log  10  =  log  3,     or    x  =  log  3  =  0 . 47712L 

4.  5'  =  |; 
taking  the  logarithms  of  both  members, 

*  log  5  =  log  f  =  log  2  -  log  3  ; 

log  2-log  3  _-.  176091 
log  5  .698970  ~ 

THEORY   OF  EQUATIONS. 

2.  Two  roots  of  the  equation, 

x*  —  I2x3  +  48*2  -  68*  +  15  =  0, 
are  3  and  5  :  what  does  the  equation  become  when  freed  of  them  t 

«*  —  12*3  +  48a;2  -  68*  +  15  \x  —  3 

x*  —  9*a  +  21a?  —  5 

xi  _    9a;2  +  21*   —  5  \x   —  5 

x2  —  4x  -f  1  =  0.     Ans. 

3.  A  root  of  the  equation, 

x*-Gxz  +  lhc-6  =  0, 
is  1  :  what  is  the  reduced  equation  ? 

xs  —  6*a  +  lla?  —  6  \x  —  1 

ara  —  5*  +  6  =0.     Ans 

4.  Two  roots  of  the  equation, 

4*<  -  14z3  -  5z2  -f  31z  -f  6  =  0, 
are  2  and  3  ;  find  the  reduced  equation. 


336-337-340.]    GREATEST  COMMON  DIVISOB.  127 

4z«  -  14z3  —  5z2  +  Six  4-  6|s-2  

4ar3  —  6x2  —  17s  —  3, 

4*3  -  6z2  —  17*  -  3  |g  — 3 

4z2  4-  6z  4-  1  =  0.     Ans 

FORMATION  OF  EQUATIONS. 

2.  What  is  the  equation  whose  roots  are  1,  2,  and  —  3  ? 

(x  -  1)  (x  -  2)  (x  4-  3)  =  z3  -  Ix  4-  6  =  0.     ^4ns. 

3.  What  is  the  equation  whose  roots  are  3,   —  4,    2  +  -y/3J  and 
2-  V§1 

(a:  -  3)  (x  +  4)  (*  -  2  -  -v/3~)  (*  —  2  4-  V§) 
=  x*  —  3x3  —  15z2  4-  49a;  —  12  =  0.     Ans. 

4.  What  is  the  equation  whose  roots  are   3  4-  \/^»  3  —  ^/dj 
and  —  6  ? 

(x  _  3  --v/5)  (a;  -  3  4-  \/5)  (a;  4-  6)  =  x3  —  32z  4-  24  =  0.    Ans. 

5.  What  is  the  equation  whose  roots  are   1,  —  2,  3,  —  4,  5, 
and  —  6  ? 

(x  -  1)  (x  4-  2)  (x  -  3)  (x  4-  4)  (x  -  5)  •(*  +  6) 
=  x6  4-  3z5  —  41**  *-  87a;3  +  400z2  4-  444a-  —  720  =  0.     Ans. 
6.  What  is  the  equation  whose   roots   are  ....  2  4- 
2—  -/"^l,  and  -  3? 

(aj  _  2  -  •v/1^!)  (a?  -  2  4-  -/^T)  (*  +  3) 
=  a;3  —  x2  —  7z  4-  15  =  0.     Ans. 

TRANSFORMATION  OP  EQUATIONS. 
2.  Transform  the  equation, 

X*  +  llx  4-  28  =  0, 
into  one  whose  roots  are  three  times  as  great. 


128  KEY  TO  DAVIES'   BOURDON.  [340-341. 


Make    z  =  -;    then  we  have, 

o 


28  =  0;    or,    ^  +  83y  +  263  =  0. 

3.  Transform  the  equation, 

^  +  3a^  —  4^  +  2^  —  a;  +  4  =  0, 

into  one  whose  roots  are  equal  to  those  of  the  given  equa- 
tion with  their  signs  changed. 
Making    x  =  —  y,    and  dividing  by     —  1, 

t/5  —  3#4  —  4#3  —  f  —  y  —  4  =  0. 
1.  Transform  the  equation, 


into  one  whose  roots  shall  be  the  reciprocals  of  those  of  the 
given  equation. 

Making    x  =  -  ,    substituting,  and  reducing, 
y*  +  3y3  +  9y  +  3  =  0. 

2.  Transform  the  equation, 

a*  +  z3  +  3x  +  2  =  0, 

into  one  whose  roots  shall  be  the  reciprocals  of  those  of  the 
given  equation. 

Making    x  =  -,    substituting,  and  reducing, 

»•+]»•  +i»+  1  =«. 

3.  Transform  the  equation, 

7         7 


into  one  whose  roots  shall  be  the  reciprocals  of  those  of  the 
given  equation. 

Making    x  =  -,    substituting,  and  reducing, 
y 


347-348.]          TRANSFORMATION    OF    EQUATIONS.  129 


TRANSFORMATION   OF   EQUATIONS   BY    SYNTHETICAL 
DIVISION. 

4.  Find  the  equation  whose  roots  shall  be  less  by  3  than 
the  roots  of  the  equation 


1  _  3  —  15  +  49  —  12  I 
-|_  3  +    0  —  45  +  12 
+  0  —  15+    4,+    0 
+  3  +    9  —  18 
+  3—    6,- 14 
+  3  +  18 


+  3 

+  9    .-.    y*  + 


5.  Find  the  equation  whose  roots  shall  be  less  by  10  than 
the  roots  of  the  equation 

z4  +  2s3  +  3s?  +  4z  —  12340  =  0. 

1+2+      3+        4-  12340  [|  10 
+  10  +  120  +  1230  +  12340 
+  12 +  123  + 1234, +  0 
+  10  +  220  +  3430 
+  22 +  343, +  4664 
+  10  +  320 
+  32, +  663 
+  10 


1,+  42    . •.    y*  +  42^  +  663y2  +  4664#  =  0. 


KEY    TO    DA  VIES5    BOURDON".  [348-349. 

6.  Find  the  equation  whose  roots  shall  be  less  by  2  than 
the  roots  of  the  equation 


1  +  0+    2  —    6  —  10  +  0  [[2- 
+  2  +    4  +  12  +  12  +  4 
+  2+    6+    6+    2,+  4 
+  2+    8  +  28  +  68 
+  4  +  14  +  34,+  70 
+  2  +  12  +  52 
+  6  +  26,+  86 
+  2  +  16 
+  8,+  42 
+  2 

1,+  10;  .-.  #5+10y+42#3+86y2+70#+4=0. 


DISAPPEARANCE  OF  SECOND  TERM. 
2.  Transform  the  equation 

7z2  +  4z  —  9  =  0 


into  an  equation  in  which  the  second  term  shall  be  wanting. 
Here  we  make  the  roots  of  the  resulting  equation  greater 

p 

than  those  of  the  given  equation  by    —  — ,    or    —2    (Bour- 

5 

don,  Art.  266) ;  that  is,  we  make  them  less  than  those  of  the 
given  equation  by     +  2.    Hence, 

OPERATION. 

1  _  10  +    7+    0+    4—    9  [2_ 
+    2-16-18  —  38-64 


349.]  TRANSFORMATION    OF    EQUATIONS.  131 

1st  quotient,        1  —  8  —    9  —    18  —   32,—  73     1st  rem. 
+  2  —  12—  42  —  120 


2d  quotient,         1  —  6  —  21—    60,—  152     2d  rem. 

+  2  —    8  —    58 
3d  quotient,         1—4  —  29,—  118     3d  rem. 

+  2-4 
4th  quotient,       1  —  2,—  33     4th  rem, 

+  2. 
5th  quotient,       1,+  0     5th  rem. 

Hence,  the  transformed  equation  is 

y*  _  33^3  _  118y8  _  152y  _  73  _  0> 

3.  Transform  the  equation 

a?  _  QXZ  +  yx  _  10  =  o 

into  one  whose  second  term  shall  be  wanting. 

Here  we  make  the  roots  of  the  resulting  equation  greater 

p 

than  those  of  the  given  equation  by     —  —  =  —  2  ;    that  is, 

o 

we  make  them  less  than  those  of  the  given  equation  by  +  2. 

OPERATION. 

l_6  +  7-  10  [2 
+  2_8—   2 


1st  quotient,        1  —  4  —  1,—  12    1st  rem. 

+  2-4 
2d  quotient,         1  —  2,—  5    2d  rem. 

+  2 
3d  quotient,         1,     0    3d  rem. 


132  KEY    TO    DAVIES'    BOITKDON.  [349. 

Hence,  the  transformed  equation  is 
ys  _  5y  _  12  —  o. 

4.  Transform  the  equation, 

a^  +  gz8  —  3  +  4  =  0 

into  one  whose  second  term  shall  be  wanting. 
Here  we  make  the  roots  of  the  resulting  equation  greater 

p 

than  those  of  the  given  equation  by     +  —  =  3 ;    that  is,  we 

o 

make   them  greater  than  those  of  the  given  equation  by  3. 
Therefore,  the  synthetical  divisor  is     —  3. 

OPERATION. 

1  +  9  —    i  +    4|_— 3 

—  3  —  18  +  57 

1st  quotient,        1  +  6  —  19,+  61    1st  rem. 
1  —  3—    9 


2d  quotient,         1  +  3,—  28    3d  rem. 
-3 


3d  quotient,        1,  +  0    3d  rem. 
Hence,  the  transformed  equation  is 
y&  _  %sy  +  61  =±  0. 

5.  Transform  the  equation 

&  _  8z3  +  7s2  +  3x  +  4  =  0 

into  one  whose  second  term  shall  be  wanting. 

Here  we  make  the  roots  of  the  resulting  equation  greater 

p 
than  those  of  the  given  equation  by     —  —  =  —  2 ;    that  is, 

wo  make  them  less  than  those  of  the  given  equation  by  +  2 ; 
hence,  the  synthetical  divisor  is     +  2. 


249-259.] 


EQUAL    ROOTS. 


133 


OPEBATION. 

1-8+    7+    3+    4  [2 

+  2  —  12  —  10  —  14 
1st  quotient,        1  —  6  —   5  -•   7,—  10    1st  rem. 

+  2—   8-26 
2d  quotient,         1  —  4  —  13,—  33    2d  rem. 

+  2-    4 
3d  quotient,         1  —  2,—  17    3d  rem. 

+  2 
4th  quotient,        1,+  0    4th  rem. 

.-.    y1  —  17y2  —  33#  —  ] 


EQUAL  ROOTS. 
4.  What  are  the  equal  factors  of  the  equation 


8s  -f  36  =  0? 
The  first  derived  polynomial  is 

7*«  -  42s5  +  50.r4  +  88z3  -  129*2  -  70*  +  48, 

and  the  common  divisor  between  it  and  the  first  member  of  the 
given  equation,  is 


The  equation 

x*  —  3z3  +  Sx2  +  7x  -f-  6  =  0, 

cannot  be  solved  directly,  but  by  applying  to  it  the  method  of  equal 
roots  ;  that  is,  by  seeking  for  a  common  divisor  between  its  first 
member  and  its  derived  polynomial, 

4z3  -  9z2  -  Ox  +  7, 

we  find  such  divisor  to  be  x  -f-  1  ;  hence,  x  -\-  1  is  twice  a  factor  of 
the  first  derived  polynomial,  and  three  times  a  factor  of  the  fir?t 
member  of  the  given  equation  (Art.  271). 


134  KEY  TO  DAVIES'  BOURDON.  [359. 

Dividing, 


6  =  0,    by     (a  +•!)*  =  *2  +  2ar  +  1, 
we  have,  x2  —  bx  +  6, 

which  being  placed  equal  to  0,  gives  the  two  roots 

x  =.  2         and         x  =  3, 
and  the  two  factors,   x  —  2         and         x  —  3. 

Therefore,  (a:  —  2)  and  (#  —  3),  each  enters  twice  as  a  factor  of  the 
given  equation  ;  hence,  the  factors  are 

(x  -  2)2  (x  —  3)2  (x  +  I)3.     Ans. 

5.  What  are  the  equal  factors  of  the  equation 

—  33**  +  27z  -  9  =  0  ? 


The  first  derived  polynomial  is 

7*6  -  18*5  +  45z4  -  76z3  +  81*2  -  66*  +  27, 

and  the  common  divisor  between  it  and  the  first  member  of  the 
given  equation,  is 

x*  —  2x3  +  4z2  —  Qx  +  3. 

The  equation 

x*  —  2x3  +  4z2  —  Gar  +  3, 

cannot  be  solved  directly,  but  by  applying  to  it  the  method  of  equal 
roots,  as  in  the  last  example,  we  find  the  derived  polynomial  to  be 

4x3  —  6x2  +  80;  —  6 

and  the  common  divisor  to  be  x  —  1  ;  hence,  (x  —  1)  enters  twice 
as  :i  factor  into  the  derived  polynomial,  and  three  times  as  a  factor 
into  the  first  member  of  the  given  equation. 
Dividing 

xt  —  2a;  f  4z2  —  6*  +  3     by     (x  —  I)2  =  a;2  —  2x  +  1, 


359-375-384.]     SMALLEST  LIMIT  IN   ENTIRE   ROOTS.       135 

we  have  for  a  quotient        xz  +  3 ; 

hence,     (z2  +  3)    enters  twice  as  a  factor  into  the  first  mem- 
ber of  the  given  equation ;  hence  the  factors  are 
(x  -  I)3  (x  +  3)3. 

SUPERIOR  LIMIT  IN  ENTIRE  ROOTS. 
2    What  is  the  superior  limit  of  the  positive  roots  in  the  equation 

xs  _  3^4  _  QX3  _  25z2  +  4z  —  39  =  0  ? 
Recollecting  that  if  we  use  x  for  x'  (Art.  285),  we  have 
X  =    x5  —  3z*    —  8z3   —  25z2  +  4z  —  39 
T  =  5z4  -  12z3  -  24z2  -  50*  +  4, 

Z  =  20x3-  36a;2  -  48*  -  50, 

V=  60s2  -  72z  -  48, 
TF=  120*  -  72. 

r=120 

The  least  whole  number  that  will  render  all  these  derived  polyno- 
mials positive  is  6  ;  hence,  6  is  the  superior  limit. 

3.  What  is  the  superior  limit  of  the  positive  roots  in  the  equation 
x5  —  5z4  —  13*3  —  17x2  -  69  =  0. 

The  process  of  solution  is  the  same  as  in  the  last  example,  and  the 
limit  is  found  to  be  7. 

COMMENSURABLE    BOOTS. 

2.  What  are  the  entire  roots  of  the  equation 
x*  —  5x3  +  25*  —  21  =  0  ? 

The  divisors  of  the  last  term  are  +  1,  —  I>  +3,   —  3,  -|-  7,  —  7 
+  21,  and   -21:  L  =  22;   -  L"=  -  4. 


KEY  TO  DAVIES'  BOURDON.  [384. 

+  21,     +7,     -f-3,  +1,  -1,     -3,     - 

-    1,     -  3,     -7,  -  21,  +  21,  +  7,     + 

+  24,     4-  22,  4-  18,  4-    4,  4-  46,  +  32,  + 

4-    3,  4-    G,  4-    4,  -46, 

4-    2,  4-    4,  +  46, 

•3,  -    1,  +  41, 

-     1,  -       1;  -41, 

therefore,    -}-  3  and   -f  1   are  the  two  entire  roots.     Dividing  the 
first  member  of  the  equation  by  the  product  of  the  factors 

(x  -  3)  (x  -  1)  =  x>  -  4x  +  3, 
we  have  x2  —  «  —  y  =  0. 

NOTE. — In  the  4th  line  we  add  the  co-efficient  of  x*,  which  is  0,  and  then 
divide  by  the  divisors,  and  thus  obtain  the  6th  line. 

3.  What  are  the  entire  roots  of  the  equation 

15z5  -  19*4  +  6a;3  +  15*2  -  I9x  +  6  =  0  ? 
+    6,     +    3,     +    2,     4-    1,     -    1,     -    2,     -    3,     -    6, 
4-    1,     4-    2,     4-    3,     4-    6,        •    6,        -3,  2,     -    1, 

-18,     -17,     -16,      -13.      -25,      -22,     -21,     -20, 
-    3,  -    8,      -13,      t-25,     +  11,     +    7, 

+  12,  +    7,     +    2,     +  40,     +  26,     +  22, 

4-2,  4-2,     -40,     -13, 

+    8,  +8,     -34,     -    7, 

+    8,     4-34, 
-11,     4-15, 
-11,      -15, 
4-  15: 
hcuoe,  there  is  but  one  entire  root,  which  is   --  1 


384-392.]  NUMBER    OF    REAL    ROOTS.  137 

4.  What  are  the  entire  roots  of  the  equation 

9z«  +  30s5  +  22Z4  +  10s8  +  17s2  —  20z  +  4  =  0  ? 

This  is  worked  like  the  preceding  example,  giving  the  en- 
tire root,  —  2.  Then  dividing  the  equation  by  x  +  2,  we 
find  a  new  one,  which  has  a  root,  —  2. 

NUMBER  OF  REAL  ROOTS. 

3.  What  is  the  number  of  real  roots  of  the  equation 

a?_5z2_|_  8z  —  1=0? 

By  finding  the  expressions  which  indicate,  by  their  change  of 
sign,  the  existence  of  real  roots  (Art.  293  and  Example  1),  we 

have 

X=    z3—    5y?+Sz  —  I 

X,=  3Z2  —  IQx  +  8 
X8=  2x  —  31 
X8=  —  2295 

x  =  —  oo    gives 1 2  variations, 

x  =  +  oo    gives     +  +  H 1  variation ; 

hence,  there  is  one  real  and  two  imaginary  roots  (Art.  293). 

For    x  =  0,    we  have 1 2  variations, 

for     x  =  1,         "  +  H 1  variation ; 

hence,  the  real  root  lies  between  0  and  +  1. 

4.  Find  the  number,  places,  and  limits  of  the  real  roots  of 

z*  —  Sz3  +  14s2  +  Ix  —  8  =  0. 
For  solution,  see  Example  3,  page  141  of  Key. 

5.  Find  the  number,  places,  and  limits  of  the  real  roots  of 
the  equation 

a?  _  23x  —  24  =  0. 


KEY    TO    DAVIES'    BOURDON.  [393-396. 

X  =  y?  —  23x  —  24 
X1=  3z2  —  23 

X2=  23z  +  36 
X3=  8279. 

For    x  —  —  oo,    we  have,     —  H 1-,    3  variations. 

For    x=  +  oo,    we  have,     +  +  +  +,    no  variations. 
Hence,  there  are  3  real  roots,  which  are  easily  placed. 

6.  In  the  sixth  example,  we  have, 

X  =  y?  +  f z2  —  2x  —  5 
X1=  3z2  +  3x  —  2 
X2=  llx  +  28 
X3=  -  1186. 

Hence,  there  is  but  one  real  root,  and  that  lies  between 
the  limits  1  and  2. 


CUBIC  EQUATIONS. 

1.  What  are  the  roots  of  the  equation 

X3  _  6z2  +  3*  =  18     .     •     .     (I)1? 

Transforming  so  as  to  make  the  second  term  disappear, 
z3- 9* -28  =  0     •     •     •     (2) 
^=  —  9         q  =  —  28  ; 
substituting  in  Cardan's  formula,  and  reducing, 

*  =  4. 

But  the  roots  of  the  given  equation  are  greater  than  those  of  equa- 
tion (2)  by  2 ;  hence  x  =  64 


396-405.]  CUBIC  EQUATIONS.  139 

Transposing   18   in  equation  (1),  and  dividing   both  numbers  by 
x  —  6,  we  find 

a:2 +  3  =  0        . ' .         *  =  ±  v/—  3; 

hence  the  three  roots  are         6,     -y/  —  3,     —  ^/  —  3. 

2.  What  are  the  roots  of  the  equation 

X3  _  9^2  +  28z  =  30     ....     (1)? 
Transforming,  we  find 

x3  +  x  —  0  (2)     .'.     x=0     and     a;  =  ±  v'~  1. 
But  the  roots  of  (1)  are  greater  than  those  of  (2)  by  8; 

hence  the  roots  of  (1)  are,     3,  3  +  T/  —  1     and    3  —  y'  —  1. 

3.  z3  —  7z  +  14  =  20 (1) 

Transforming  (Bourdon,  Art.  266),  we  have, 

7         344  7  344 

3*~T~         ^'  6'    ^=~~3'    9'=~T7~4 

g 

Substituting  in  Cardan's  formula,  we  have  ^ ,  the  real  root. 

o 

7 
But  the  roots  of  (1)  are  greater  by  -  than  those  of  (2) ;  hence, 

B 
in  (1)  x  =  5. 

Transposing  and  dividing  by    x  —  5,    we  have, 
a;2  _  2x  +  4  =  0 ; 

hence,  the  required  roots  are,  5,  1  +  V—3,  and  1  —  V—  & 

HORNER'S  METHOD  OF  SOLVING  NUMERICAL  EQUATIONS. 
1.  3?  +  z2  +  x  —  100  =  0. 

By  Sturm's  Rule,  we  find 


140 


KEY  TO  DA  VIES5  BOURDON. 


[405. 


X    r=  X3  +  XZ  +  X     -  100, 

Zl  =       3*2  -f  2*  +  1 
Xz  =  -  4x  +  899 
X,  =  -  2409336. 

For        cc  =  —  oo ,  1--| , 

for          *  =  +  CD  ,  -f  H , 

hence,  there  is  but  one  real  root. 
Tor         x  =  4,  .  +  +  .-, 

for  x  =  5,  +  +  +  _  , 

hence,  the  real  root  lies  between  4  and  5. 

2.  a*  -  12*2  4-  12*  -  3  =  3. 

By  Sturm's  Rule, 

X  =x*    -  12*2  +  12* -3, 

JTj  =  4*3  -  24*  +  12,     or    x3  -  Qx  +  3, 


2  variations, 

1  variation  ; 

2  variations, 
1  variation ; 


X,  =  13*-    9, 

X4  =  20. 

For  x  —  —  oo ,  +  —  - 

for  x  =  +  oo  ,  H — \-  • 
hence,  there  are  4  real  roots. 

For  x  =  —  4  H f- 

for  *  =  —  3  \- 

for  *  =       0  h  + 

for  x  =  -f  1  =F 

for  .T  =  +  2  h 

for  *  =  +  3  +  +  4- 


4  variations, 
0  variation ; 

4  variations, 
3  variations, 
3  variations, 
1  variation, 
1  variation, 
0  variation ; 


406.]  HORNER'S  METHOD.  141 

hence,  one  of  the  roots  lies  between  —  4  and  —  3,  two  between  0 
and  1,  and  the  remaining  root  lies  between  2  and  3. 

3.  x*  -  8*3  +  14*2  +  4z  -  8  =  0. 

By  Sturm's  Eule, 

X  =  x*    —  Sx3    +  14*2  +  4*  —  8, 

XI  =  4*3  —  24*2  +  28*  +  4,     or    x?  —  6*2  -f  7*  +  1, 
X2  =  5z2  -  17*  -j-  6, 

X3  =  76*  -  103, 
X<  =  45475. 

For     x  =  —  co  ,  H 1 (•  4  variations, 

for      £=  +  00,  +  +  +  +  +  0  variation ; 

hence,  the  equation  has  4  real  roots. 

For     *  =  —  1  + h 1-  4  variations. 

for*=       0  h-l h  3  variations, 

for      orrr-fl  +  + f-  2  variations, 

for      x  =  +  2  H 1 — H  2  variations, 

for      z=:-f3  —  —  ±  +  +  1  variation, 

fora;=       5  H  +  +  1  variation, 

fora;=       6  +  +  +  +  +  0  variation ; 

hence,  one  root  is  between  —  1  and  0,  one  between  0  and  1,  one 

between  2  and  3,  and  one  between  5  and  6. 

4.  x5  —  I0x3  +  Qx  +  1  —  0. 
By  Sturm's  Rule, 

X  =  x5      -  10*3+  6*  +  1, 

XI  -  5*4  -  30*2+  6, 
X2  =  20*3  -  24*  -  5, 
X3  =  96*2  -  5*  -  24, 
X4  =  43651*  +  10920, 
X,,  =  32335636224. 


142 


KEY  TO  DAVIES'  BOURDON. 


[406. 


For    x  =  ~  oo         1__. I-  __'_!_ 

for      ar=+ao          -f.  +  _}.  _j_  _j_  _j_ 

hence,  the  equation  has  5  real  roots. 


For    x  =  —  4 


for 
for 
for 
for 
for 
for 

hence, 

ar=-3 

x=  —  I 

x  =      0 
*  =  +  ! 
x-  +3 
x  =.  +  4 

one  roc 

i      i  i 

+    4-    -   +  ~    + 

4- 4-  ~    + 

+    +    --  +    + 


4-  4-    4-    + 

+    +    +  +    +    + 

one  root  lies  between  —  4    and 

two  roots  lie  between  —  1  and 
one  root  lies  between  0  and 
one  root  lies  between  3  and 


5  variations, 
0  variation ; 

5  variations, 
4  variations, 
4  variations, 
2  variations, 
1  variation, 
1  variation ; 
0  variation ; 

-3, 

0, 

4-1, 

4 


APPENDIX. 


GENERAL  SOLUTh,*  OF  TWO  SIMULTANEOUS  EQUATIONS  OF  THE 

FIRST  DEGREE. 
1.  Take  the  equations, 

ax  +  by  —  c      .     .     .     (1), 
a'z+b'y  =  c'     ...     (2); 

multiply  both  members  of  (1)  by  b'  and  of  (2)  by  6,  then  sub- 
tracting and  factoring,  we  find 

(aV  -  a'b)  x^b'c-  be' ; 
b'c  -  be' 


ab'  -  a'b 


(3). 


r    vi       '  ac'  —  a'c 

In  like  manner,  y  =  — 77     •     •     •    (4). 

ab  —  ab 

By  means  of  formulas  (3)  and  (4)  any  two  simultaneous  equations 
of  the  forms  (1)  and  (2)  may  be  solved. 

Thus,  4x  -f  By  =  31, 

3*  +  2y  =  22  : 
by  comparison  with  (1)  and  (2), 

a  =  4,     6  =  3,     c  =  31,    a' =  3,     b'  =  2,    c'  =  22; 
by  substitution  in  (3)  and  (4), 

62  -  66  88  -  93 


144  APPENDIX. 

EXAMPLES. 

(     -+y-   =2     ) 

1.  Given  J     3      4  I     to  find  x  and  y. 

I  3x  +  4y  =  25  ) 
By  comparison  with  (1)  and  (2), 

a  =  £,         6=1,         c  =    2, 

a'  =  3,         ^  =  4,         c'  =  25  ; 
by  substitution  in  (3)  and  (4), 

X  =  *L--J%  =  8>  y   =  8^~6   =   4. 

**•  ~~      4  *3   ~~    4 

2.  Given  4  f-      to  find  a;  and  y : 

(-  5z    +  16y  =  124) 

by  comparison, 

a  =       11,         b  =  —    5,         c  =  —  1, 
a'  --=  -    5,         b'  =        16,         c'  =        124 ; 

oy  substitution, 

-  16    +  620  _  1364  —  5 

176-25  '  176  -  25  ~ 

GENERAL  SOLUTION  OF  THREE  SIMULTANEOUS  EQUATIONS  OF 
THE  FIRST  DEGREE. 

2.  Take  the  equations, 

ax    +  by    +  cz    =  d     •     •     •     (1), 
a'x  +  Vy  +  c't  =  d'    •     •     .     (2), 

~n~,    i    irr.,    i     ..//_        j//  /o\ 

ax-}-by  +  cz-=a     •     •     •     (6). 

From  (1)  and  (2)  we  obtain,  by  eliminating  e, 

(So,  -  ca'}  x  +  (c'b  -  cb')  y  =  c'd  -cd/     •     -     -     (4). 


ADDITIONAL   EXAMPLES.  145 

In  like  manner,  from  (1)  and  (3), 

(c"a  -  ca")  x  +  (c"b  -  cb")  y  =  c"d  -  cd"         •     •     •     (5;  ; 
combining  (4)  and  (5)  and  eliminating  y,  we  find 

_  (c"b  -  cb")  (c'd  -  cd1)  -  (c'b  -  cb')  (c"d  -  cd") 
~  (c'a  —  ca' )  (c"b  -cb")  -  (c"a-  ca")  (c'b  -  cb') 

In  like  manner, 

_  (c'a  -  cap  (c"d  -  cd")  -  (c"a-  ca")  (c'd  -  cd') 
y  -  (c'a  _  ca')  (c"b  _  cb")  -  (c"a-  ca")  (c'b  -  cb')  *** 

_  (a"6  -  ab")  (afd  -  ad')  -  (a'b  —  ab')  (a"d  —  ad") 
~  (c'a  -  ca')  (b"a  -ba")  -  (c"a-  a"c)   (b'a  -  ba' ) 

Formulas  (6),  (7)  and  (8)  enable  us  to  solve  all  groups  of  simul« 
taneous  equations  of  the  form  of  (1),  (2)  and  (3).     Thus, 

2x  +  Sy  +  4z  =  29, 
3*  +  2y  +  50  =  32, 
4x  +  3y  +  2«  =  25 : 
Vy  comparison  with  (1),  (2)  and  (3), 

a    =2,        b    =3,        c    =4,        d    =29, 
a'  =3,        b'  =  2,        c'  =5,        <T  =  32, 
a"  =  4,         b"  =3,         c"  =  2,         <*"  =  25: 
by  substitution  in  (6),  (7)  and  (8), 

_  (  6  -  12)  (145  -  128)  -  (15  -    8)  (58  -  100) 
~  (10  -  12)  (     6  -  12  )  -  (  4  -  16)  (15  -  8) 

-  102  +  294  _  192  _ 

:  =     ~      ' 


146 


APPENDIX. 


-  -10  -  *2)  (58  -  100)  -  (4  -  16)  (145  -  128) 
~  (10  -  12)  (~6  -  12  )  -  (4  -    6)  (~15~^  8     ) 

84  +  204  _  288  _ 
=  12  +  84    :  "  "96"  =      ' 

_  (12-    6)  (87  -  64)  -  (9  -    4)  (116  -  50) 
~  (10  -  12)  (  6  -  12)  -  (4  -  16)  (     4  -  9  ) 

138  -  330  _  192  _ 

~  +12  —  60    : "  "48~  Z 


1.  Given 


EXAMPLES. 

x  +    y  +  z  =       90 
x  —  3y         =  —  20 
I  2*  -  4z         =  —  30 


By  comparison  with  (1),  (2)  and  (3), 

a  =  1,  b  =  1,  c  =  1, 
a'  =  2,  b'  =  -  3,  c'  =  0, 
a"  =  2,  b"  =  0,  c"  =  -  4. 

by  substitution, 

-  4  X  20  -  3  x  -  330      910 


to  find  x,  y  and  z. 


d  =  90 
df  =  -  20, 
d"  -  -  30 ; 


«•  — 


—  2x  -4  +  6x3          "26 

—  2  X  —  330  +  6  X  20  _  780  _ 
-2x4  +  6x3  =  "26~  "        ' 


z  — 


2  x  200 


5  x  210     _  —  650  _ 


_'2X-2+6x-5        —  26 


2.  Given 


x+    y  +    z=    61 
x  -f  2y  +  3s  =  14 
3ar  _  y    +  4s  =  13  J 


to  find  ar,  y  and 


ADDITIONAL   EXAMPLES. 


147 


by  comparison, 

a    =1, 

b 

=       1, 

c    =1, 

d    =    6, 

a'   =1, 

b 

'  =       2, 

c'  =3, 

dx  =  14, 

a"  -  3, 

b 

"  —        i 

J5 

c"  =  4, 

d"  =  13  : 

by  substitution, 

5x4-11 
*  —  —  1  • 

11 

2x11—4 

-2-       s  — 

4X-8+1XE 

2x5-1 


2x    5-1 


2X-4-1 


=  3 


ELIMINATION  BY  THE  METHOD  OF  ARBITRARY  MULTIPLIERS. 

3.  There  is  a  method  of  elimination  by  means  of  arbitrary  quan- 
tities that  will  often  be  found  useful,  particularly  in  the  higher 
investigations  of  applied  mathematics.  It  consists  in  multiplying 
both  members  of  one  of  the  given  equations  by  an  arbitrary  quan- 
tity, then  adding  the  resulting  equation  to  the  second  of  the  given 
equations,  member  to  member,  after  which  such  a  value  is  to  be 
assigned  to  the  arbitrary  quantity  as  will  reduce  the  co-efficient  of 
the  quantity  to  be  eliminated  to  0. 

To  illustrate,  let  us  take  the  two  simultaneous  equations, 

ax  +  by   —  c      ...     (1), 
a'x  +  b'y  =  c'     -     •     -     (2) ; 

multiplying  both  members  of  (1)  by  n,  which  is  entirely  arbitrary, 
we  have 

nax  +  nby  =  nc     •     •     •     (3) ; 

adding  (2)  and  (3),  member  to  member,  and  factoring, 

(na  +  a')  x  +  (nb  +  b')  y  =  nc  +  c'     •     •     .     (4). 
If  it  be  required  to  eliminate  y,  place 


148 


APPENDIX. 


substituting  this  in  (4)  and  reducing,  we  find 
~~°  +  C'      b'c-bc' 


If  it  be  required  to  eliminate  x,  place 

no,  +  a'  =  0  ;         .'. 
substituting  in  (4)  and  reducing,  we  find 


-c  +  c' 

a  etc'  —  a'c 


-?Lb  +  b>       ab'~a'b 
a 

These  values  of  x  and  y  correspond  to  those  already  deduced  by 
previous  methods. 

As  an  example,  let  it  be  required  to  find  the  values  of  x  and  y 
from  the  equations 

7z  +  3y  =  33     •     •     •     (2)  ; 
multiplying  both  members,  of  (1)  by  n, 

3nx  —  ny  =  5ra     •     •     •     (3)  ; 
adding  (2)  and  (3),  member  to  member, 

(3%  +  7)  x  +  (3  —  n)  y  —  5n  +  33    •     •     •     (4) : 
1  st.  Assume        3  —  n  =  0 ;        .  • .     n  =:  3 ; 
substituting  in  (4),  and  reducing, 

15  +  33  _  Q 

:    9  +  7 

2d.' Assume        3n  +  7  =  0; 


ADDITIONAL   EXAMPLES.  140 

substituting  in  (4),  and  reducing, 

=  4. 


S+3 


EXAMPLES. 


1.  Given  •{  }•      to  find  x  and  y. 

V,  J 

Multiplying  both  members  of  the  first  equation  by  n  and  adding  to  the 
second,  member  by  member, 


making  n  =  —  -        and  reducing, 

m 


making  n  =  -        and  reducing, 


2   Given 


y-2 

/>• __,_-.  „•  —  ^ 

a?          7        -o 


Q 
4y —  =3 


to  find  x  and  y. 


Reducing  and  transposing, 

7*-y  =  33     •     •     •    (1), 
12y  -  x  =  19     •     •     •    (2)  : 

multiplying  by  n,  and  adding  and  factoring, 
9 


150  APPEltDIX. 

(In  —  1)  x  —  (n  —  12)  y  =  33n  +  19 : 
making  n  =  - ,      we  find      y  =  2 ; 

making  n  =  12,        we  find        x  =  5. 

T+?  =  s 

3.  Given  -^  ^      to  find  x  and  y. 

~jr~  ~7  = 
Multiplying  by  n,  adding  and  factoring, 

(9n  +  14)  -  +  (6%  -  6)  -  =  36/i  +  10  ; 

14  1  1 

making  n  = — ,     we  have    -  =  o ;       .  • .  y  =  -> 

y 

making  n  =  1,  we  have     -  =  2;       .  • .  x  =  -• 


MISCELLANEOUS  GROUPS  OF  SIMULTANEOUS  EQUATIONS  OF  THE 
FIRST  DEGREE. 

82;       5y       9 
] .  Given  <  >      to  find  a:  and  y. 

—  +  —  -  - 

i,  5a;      3y      4 

Combining  and  eliminating  -, 

tG 

OK          Of  «         Z^         T9  ' 
itii)         Jf/  y         to          is/ 

1          DO  i 20 

whence,  -  =  57 »         ?  -.  •  •  1  r=  » IT 


ADDITIONAL   EXAMPLES. 


Conr/Dining  and  eliminating  -» 
y 


\  _  JLU  -  —    — 

9      25/  x  ~  27       12 ' 


1        65 
whence,  x=192' 


=  5 


2.  Given  •{  i.    to  find  x  and  y ; 

I      r*.  * 

c  —  1 
5  J 

Clearing  of  fractions  and  reducing, 

*  +  6y=     15     •     •     •    (1), 
2z  —  5y  =  —  4     •     •     .     (2)  : 
combining  and  eliminating  ar, 

Yly  =  34 ;         .  • .     y  —  2  ; 
substituting  in  (1),  ar  =  3. 

y-2 

A.  £ 

ry      ~"       ~— 

3.  Given  \  \      to  find  a?  and  y. 

3 
Clearing  of  fractions  and  reducing, 

7a;-y  =  33    •    •    •     (1), 
12y-ar=19     •     .     .     (2); 
combining  and  eliminating  ar, 

83y  =  166;         .-.     y  =  2; 
substituting  in  (1),  x  =  5. 


152 


APPENDIX. 


4.  Given 


5*6     4  +  2y  =  24 


to  find  x  and  y. 


Clearing  of  fractions  and  reducing, 

5z  -f  12y  =  148, 
25*  —    2y  =  182  ; 
combining  and  eliminating  y, 

155*  =  1240  ;         .  • .     x  -  8  ; 
substituting,  we  find  y  =  9. 


5.  Given 


6 


to  find  x,  y  and  c. 


Clearing  of  fractions  and  reducing, 

3x  +  2y+    s  =  72    .     .     .     (1), 
—  x  +  3y  +  20  =  48    •     •     .     (2), 
3*+          2*  =  60    ...     (3); 
combining  (1)  and  (2),  eliminating  y, 

11* -2=  120     •     •     .     (4); 
combining  (3)  and  (4),  eliminating  2, 

25*  =  300  ;         .  • .    *  =  12 ; 
by  substitution  in  (3),  2  =  12, 

"         "  (1),  y=»12. 


ADDITIONAL   EXAMPLES. 


153 


6.  Given 


Clearing  of  fractions, 

21*+14y  +    60  =  924 
10s  -f    6y+152  = 
x+    2y+      2  =  11 

combining  (1)  and  (3),  eliminating  2, 


to  find  or,  y  an<    <. 


(1), 
(2), 
(3); 


(4); 
(5)  ; 


combining  (2)  and  (3),  eliminating  2, 

5*  +  24y  =  780    •     • 
combining  (4)  and  (5),  eliminating  ar, 

70y  =  2100 ;  .  • .    y  =  30 ; 

from  (5),  x  -  12 ;     from  (3),     z  =  42. 

f  2*  -  3y  +  22  =  13  •     •     •  (1) 

2v  —    *  =  15  ...  (2) 

2y+    «=    7  ...  (3) 

I  5y  +  3v  =  32  .     .     .  (4)  J 
Combining  (2)  and  (4),  eliminating  v, 

lOy  +  3x  =  19  •    •    •  (5)  ; 
combining  (5)  and  (1),  eliminating  a?, 

29y  -  62  =  -  1  .     .     .  (6)  ; 


7.  Given 


t(  find  x,  y,  t 
and  v. 


154 


APPENDIX. 


combining  (6)  and  (3),  eliminating  z, 

41y  =  41;         .'.     y  =  1  : 
by  successive  substitutions,        z  =  5,     x  =  3,     v  =  9. 


8.  Given 


3       5       19      4 
z  +  z~24  +  y 


24 


y      2  ~   8 

Transposing,  reducing,  &c., 


x          y 


a;          y         z       24 


(1), 
(2), 
(3): 


combining  (1)  and  (3),  also  (2)  and  (3),  eliminating  -, 

—  10- +43-  =  —     •     •     •     (4), 
x  «/24 

8^-11y==24     '     '     '     (5)J 

combining  and  eliminating  -, 
x 

2B*l  =  i&>     •'•  p  =  l?  or  y=12> 

substituting  in  (5),       a;  =  6 ;     whence;     2  =  8. 
f     10+  6y—  4a;  _  4 

9.  Given  <(  f-      to  find  x  and  y. 


126+  8a;-17y  _  35 
100— 12ar+"7y  ~  13  J 


ADDITIONAL   EXAMPLES.  155 

Clearing  of  fractions  and  reducing, 

-  2*  +      3y  =  -      1     ...     (1), 
262*  -  233y  =       931     •••     (2). 
Combining  and  eliminating  x, 

160y  =  800;         .'.     y  =  5; 
by  substitution  in  (1),  x  =  8. 

r  ax  +  by  =  c2  \ 

10.  Given  •<  a(a  +  x)_      >     to  find  x  and  y. 

(W+7)-      ) 

Clearing  of  fractions  and  reducing, 

ax  -f-  by  =  c2    -     «     «    (1),   ' 

ax  — by  —  ^  -  a2      •     (2)  ; 

•>y  addition, 

52  _i_  r2 a2 

2aa;  =  62       c2-a2  .-.     x  = 


2a 
by  subtraction, 

2£y  =  c2  -f-  a2  —  62 ;          .  • .     y  = — ^- 


MISCELLANEOUS  EXAMPLES  OF  EQUATIONS  OF  THE  FIRST,  SECOND 
AND  HIGHER  DEGREES,  CONTAINING  BUT  ONE  UNKNOWN 
QUANTITY. 

1.  Given  3*2  —  4  =  28  +  *2,     to  find  x : 
transposing  and  reducing, 

x2  =5  16 ;        .  • .     x  =  ±  4 

2.  Given         ^-±-5  -  £±*?  =  117  -    fo»,     to  find  x; 

o  o 


1 56  APPENDIX. 

Clearing  of  fractions,  transposing  and  reducing, 
xz  =  25  ;        .  • .     *  =  ±  5. 

3.  Given  a;2  +  ab  =  5a:2,     to  find  x ; 

transposing  and  reducing, 

*2  =  V '         '   '     *  =  ±  2 


4.  Given 


+7        a?   -7 


;o  find  a?. 


a;2  —  7*      a;2  f  7a;      a;2  —  73' 
Clearing  of  fractions, 
x*  4-  14a:3  -  24a;2  -  1022a:  —  3577  —  a;*  +  14a;3  4-  24a:2  —  10222 

4-  3577  =  7a:3  —  343a: ; 
transposing  and  reducing, 

21a;3  =  1701ar,     or    z2  =  81 ;         . '.     a?  =  ±  9. 

/*-2  .      /*  +  2 
5.  Given          W      ,   2  +  V     _  2  ~  ^>     to  fin<^  * ' 

multiplying  both  members  by  <^x  4-  2, 

-/a;  —  2 
multiplying  both  members  by  -y/a;  —  2. 

x  —  24-a;4-2  =  4  -\/xz  —  4,     or    a;  =  2  y^a;2  —  4 ; 

squaring  both  members, 

16       16  4     /— 

x»  =  4a;2  —  16,     or     a;2  =  —  =  —  X  3  ;         . ' .     a;  =  ±  -  -/3. 


6.  Given 


x  4-  y^Sa;  4-  10  =  8,         .o  find  x ; 


ADDITIONAL   EXAMPLES. 


Transposing  and  squaring  both  members, 
5x  +  10  =  64  —  IGz  + 
whence,  xz  —  21a;  =  —  54  ; 


21          /          441       21       15 

by  the  rule,        x  =  —  ±  \/54  +  —  :  =  —  ±  —  j 

.  • .     a;  =  3,         *  =  18. 

7.  Given  5  \f*  +  7  ^  _  108,       to  find  x : 

make  \f&  =  y  ;     whence,     ^/z*  =  y2 ; 

substituting  and  reducing, 

7          108 


7          /108       49  '  7       4? 

whence,  y=    -  JQ  *  \/5    'f  100  =    ~TO±10; 

27 

.  • .     y  =  4     and    y  = —  ; 

5 


27\3 


from  which,     x=±^y3=±S    and    a;=±y/y^=±  \/| — ^- 

8.  Given  3z2  +  10z  =  57,       to  find  x. 

By  division, 


+  — -  x  =  19  ;     whence, 

o 

_5       14 

/.     x  =  3    and    a:  =  —  6^« 

9.  Given  (a;  -  1)  (x  —  2)  =  1,     to  find  x. 

Performing  indicated  operations  and  reducing, 

o;3 Qj.  —    1   . 

•*'          *  «J^/      —       ^~*      X      • 


158  APPENDIX. 


115 

10.  Given  -  z2  —  -  x  =  -,     to  find  a:. 


Dividing  both  members  by  -,  or  multiplying  by  2, 

m 

2          5 

xz  —  -  x  =  -  ;     whence, 

_  5       1       1       7 

- 


.-.     x  =11,     *=-£ 

11.  Given  *LZli?_i±*       a,     to  find  *. 

8  —  x        x  —  2 

Clearing  of  fractions, 

2z2  —  14z  +  20  —  (5x  +  24  —  x2)  =  20s  —  32 

,    .  39  28 

reducing,  a;2  —  —  -  x  =  --  —  ;     whence, 

•  •  u 


39         /     28  ,    1521       39  ±  31 

X  —  •— r  ±\/ h     ,„„    =  

10      V        5         100  10 

4 


12.  Given  __  -  __  =      ,    to  find  A 

Clearing  of  fractions, 

35  (x  +  3  —  a;  +  1)  =  &  -\-  2a;  —  3 ; 
reducing,  xz  +  (2x  =  143  ;     whence, 

x  —  -  1  ±  -v/l4T=:  -  1  ±  12; 


ADDITIONAL   EXAMPLES.  150 

.-.     *  =  11,         *=  -13. 

24 

13.  Given  x  H  --  -  =  3r  —  4,    to  find  x. 

x  —  1 

Clearing  of  fract  ons, 

x2  —  x  +  24  =  3*2  —  3x  —  4x  +  4; 
reducing,  x2  —  3x  =  10  ;     whence, 

3,      /in,9      3±7 

^-±^10  +  -  =  —  ; 

.  •  .     x  =  5,         *  =  —  2. 

a;  x  +  1       13 

14.  Given  —  —  -  +  -  -  =  -3-,     to  find  ar. 

a;  +  1  a;  6 

Clearing  of  fractions, 

6z2  +  Qxz  +  12a;  +  6  =  13*2  +  13ar  ; 
reducing,  x2  +  x  =  6  ;     whence, 

-  1  ±  5 

-  =  ^-; 

.  •  .     a:  =  2,         ar  =  —  3. 

a;  —  4 

15.  Given  --  —  =  x  —  8,     to  find  a; 


Since  a;  —  4  =  (^fx  —  2)   (y^+  2), 

we  have,  by  performing  indicated  operations, 

•y/x  —  2  =  x  —  8,     or     -y/aT=  ar  —  6  ; 
squaring  both  merr  bers, 

x  =  xz  —  l2a;  +  36  j 
or,  x*  —  1  3x  =  —  36  ;     whence, 


13          f~        ~l69       13       5 

=  -±^-36  +  -  =  -^-; 


160 


APPENDIX. 

.  • .     x  —  9,        x  =s  4. 


16.  Given 
Reducing, 

*  =  - 

17*2  +  19*  - 
*2+19* 

-  1848  = 
1848 

0,     to  find  5 
whence, 

—  19  ±355 

r  17 

17    ' 

19          /I  848 

361 

34  ~V    17 

1156 

34 

17.  Given 


x  —  9  jf,    and     a;  =  —  11. 


1  5 

-  a;2  +  -  x  =  27,     to  find  ar. 

O  xi 


Multiplying  both  members  by  3, 

15 


+  —  x  =  81  ;    whence, 
.   225       —  15  ±  39 


15 


.  • .     x  =  6,     and    x  =  — 


18.  Given 


12 


-,     to  find  r. 


Transposing  and  reducing, 

28  —  a;2 


s_4-    *-4' 
squaring  both  members  and  clearing  of  fractions, 

x*  —  16  =  784  —  56z2  +  ** ; 
reducing,  **  —  57*2  =  —  800 ; 


28 


ADDITIONAL   EXAMPLES.  161 

by  Rule.  Art.  124, 

/57~       l~          ,  3249  /57      7 

x  =  db  y  —  db  y  —  800 +  —J—  =  iW-^-i-; 

hencer  x  =  ±  5,     and     x  =  ±  4-y/2. 

2*  +  9   ,   4*  -  3  ,  3a;  —  16 

19.  Given =  3  H ,    to  find  x. 

9  4x  +  3  18 

Clearing  of  fractions, 

16*2  +  84*  +  54  +  72*  -  54  =  216*  -(-  162  +  12*2  -  55*  -  48 ; 

5  114 

reducing,  *2  —  -  *  =  — -  ;     whence, 


114      »      5±48 

f  64"    ~8      ' 

.  • .     *  =  6,     and    *  =  —  4 


20.  Given        a3  +  *  +  2  -/*2  -f  a:  +  4  =  20,     to  find  *. 
Making  z2  -{-  x  =  y,        and  reducing, 

2  /y  +  4  =  20  -  y  ; 
squanng  both  members, 

4y  +  16  =  400  —  40y  +  y2 ; 
reduc'ng,  y2  —  44y  =  —  384 ;       whence, 


y  =  22  ±  V— 384  +  484  =  22  ±  10 ; 

.-.     y  =  32,     and    y  =  12  : 

taking  the  first  value  :  *  y  and  substituting  in  the  equation, 
xz  +  x  —  y, 
x2  +  x  =  32  ;     whence. 


162 


APPENDIX. 


taking  the  second  value  of  y, 

x2  +  x  =  12 ;     whence, 


21.  Giveh  \/ x 

transposing, 


squaring  both  members, 


=  x,     to  find  x. 


whence,  by  reduction, 

x*  -  x  +  1  =  2  V^ 
Placing  x2  —  x  =  y     •     •     • 


(1), 


squaring  both  members, 

yz  +  2y  +  1  =  4y ;     whence, 

y2-2y=-l;         .'.     y  =  1  ±  V-  1  +  1,    or    y= 
substituting  in  (1),         a;2  —  a:  =  1  ; 


22.  Given  a;2  —  6z  =  6a  +  28,     to  find  x : 

transposing,  a?  —  1 2x  =  28  ;     whence, 

x  =  6  ±  -y/28  +  36"=  0  ±  8  ;         .  • .     x  =  14,     «  =  —  2.  * 


ADDITIONAL   EXAMPLES.  163 

23.  Given  x**  —  2z3n  -f  x*  —  6  =  0,    to  find  « : 

making,  x*  =  y  ;     whence, 

y*  -  2y3  +  y  -  6  =  0  ; 
causing  the  second  term  to  disappear  (Arts.  263  and  313), 

3         91 

^  ~  2  Z  =  Itf ' 

id  1U 

By  the  rule  for  solving  trinomial  equations  (Art.  124). 

/3±  10 


also. 


1      . 1 

.-.     2==t--v/13,     and     z  =  ±  - 


24.  Given 


x*  -4-  0.x3   I    8 

-^_  =  ^2  +  ,,.  +  8      to  find  Xm 

x2  +  x  —  6 


Clearing  of  fractions, 
x*  -f  2x3  + 


2x3 


2*  —  48  ; 


,  2          56 

reducing,  x2  +  -  ar  =  —  ;     whence, 

O  D 


.  • .     x  =  4,     and     a;  =  —  4J. 

2 

25,  Given  *a  —  1  =  2  4-  -,    to  find  x. 

x 

Reducing,  a3  —  3a  -  2  =  0     •     •     •     (1)  ; 


164:  APPENDIX. 

comparing  with  x3  -\-  px  -f-  q  =.  0, 

P=-Z,      ?  =  -2; 

by  Cardan's  formula, 

*  =  3v/r+\A  =  2: 

dividing  both  members  of  (1)  by  x  —  2, 

xz  +  2a:  -\-  1  =  0  ;     whence, 
*=  -1, 

and  the  two  roots  are  each  equal  to  —  1  ;  hence,  the  three  roots 
»re>  +2,  —  1,  and  —  1. 

26.  Given  2<r2  -f  34  =  20*  +  2,     to  find  x. 
Transposing  and  reducing, 

xz  -  lOx  =  —  16;         .-.     x  =  8,     x  =  2. 

»  -*       £ 

27.  Given  x*  =  56*  *  +  x6,     to  find  a?. 

multiplying  both  members  by  x  ,  and  reducing, 

**  -  **  =  56  : 
comparing  with  trinomial  equations  (Art.  124),  we  find 


3          ,     1       2 
«  =  -,     and     -=_--; 


hence,  by  rule, 


I 

taking  the  upper  sign,         x  =  8    =  4, 

"    lower     "  x  =  (  —  7)    = 

28.  Given        x3  —  12z2  -f  4x  +  207  =  0,     to  find  x. 


ADDITIONAL   EXAMPLES.  165 

A  superior  limit  of  the  positive  roots  is  13  (Art.  279) ;  a.  supe- 
rior limit  of  the  negative  roots  (numerically),  is  —  7  (Art.  281). 
By  the  method  of  Art  285,  rejecting  +  1  and  —  1,  which  are 
not  roots,  we  find 

Divisors,  9,          3,         —    3 

23,        69,         -  69 
27,        73,         -  65 
3, 

-9, 
-1, 

0; 

hence,  9  is  a  commensurable  root. 

Dividing  both  members  by  x  —  9,  we  find 

3.2  _  sx  _  23  —  0 ;     or,    x2-  —  3z  =  23 ; 


*  =  r±v/23  +  ?  = 


4  2 

29.  Given  x3  +  3jr2  —  Gx  —  8  =  0,     to  find  x. 

This  is  solved  in  a  manner  similar  to  the  preceding. 

A  superior  limit  of  positive  roots  is  4,  and  of  negative  roots 
(numerically),  —  7. 

Divisors             4,            2,          1,  —  1,  -  2,  —  4, 

-2,     -    4,  -8,  +8,  +4,  +2, 

-8,      -10,  -14,  +2,  -2,  -4, 

-2,     -    5,  -14,  -2,  +1,  +1, 

+  1,        -    2,  -11,  +1,  +4,  +4, 

1,  -11,  -1,  -2,  -1, 

0,  -10,  0,  -1,          0; 

hence,  x  =  2,        x  =  —  1,        x  =  —  4. 

10 


166  APPENDIX. 

30.  Given  x3  -f  9z  —  1430  =  0,    to  find  x. 

By  the  same  rule  as  before,  we  find  14  for  a  superior  limit  of  the 
real  positive  roots,  and  from  Art.  283  we  see  that  the  equation  has 
no  real  negative  roots.  By  the  rule  (Art.  285),  we  have 

13,           11,           10,             5,             2,  1, 

-110,     -130,  -143,  -286,  -715,  -1430, 

-101,     —121,  -134,  -277,  -706,  -1421, 

-      8,     -    11,          ...,            ...,  -353,  -1421, 

...,      —      1,           ...,             ...,             ...,  —  1421, 

0,          ...,             ..,            ...,  -1420, 

hence,  1 1  is  the  only  commensurable  root. 
Dividing  both  members  by  x  —  11,  we  have 

xz  +  H*=  -  130; 


31.  Given 


Clearing  of  fractions  and  transposing, 

-V/*2  +  *lx  =  21  -*; 
squaring  both  members  and  reducing, 

49s  -  441  ;         .  •  .     x  = 


32.  Given  -/a  -f  x  —  ya  —x  =  -^/ax,     to  find  x. 

Squaring  both  members  and  reducing, 


2a  —  ax  =  2  -/a2  —  a;2 ; 
squaring  both  members, 

4a2  —  4a2*  +  a2*2  =  4a2  — 


ADDITIONAL   EXAMPLES.  167 

reducing  and  dividing  both  members  by  x, 

4a2 
4a2;         .-.     x  =  ^      ^ 


33.  Given         ^4a  +  a:  =  2  -y/6  +  *  —  V^i     ^°  ^^  r* 
Squaring  both  members  and  reducing, 

(a  —  6)  —  x  =  —  Y/ta  +  x2  ; 
squaring  both  members, 

(a  —  6)2  —  2  (a  —  6)  x  -\-  x2  -  bx  +  x2  ; 

hence,  (2a  -  6)  *  =  (a  -  6)2  ;         .-.     s  =  ^  ~  ^ 

34.  Given       -y/4a  +  a;  +  -^/a  +  a;  =  2  ^/x  —  2a,     to  find  *. 
Squaring  both  members  and  reducing, 


+  Sax  -+-  x2  =  2x  —  13a  ; 
squaring  both  members, 

16a2  -j-  20aa;  -j-  4a;2  =  4a;2  —  52aa;  -f-  169a2 ; 

reducing,  72aa?  =  153a2 ;         .  • .     x  =  — — 

8 

1  4.  2-3          1  —  z3 

35.  Given  — — — rr  +  TT ^  =  a,     to  find  x. 

(1  4-  xf       (1  —  a:)2 

Dividing  both  terms  of  the  first  fraction  by  (1  •}-  x),  and  of  the 
second  by  1  —  x,  we  have 

1  —  x  +  x2      14-a;4-*2_ 
1  4-  x  I  —x          a' 

clearing  of  fractions, 

1  —  2a;  4*  2a;2  —  x3  4~  1  4-  2a;  4"  2a;2  -{-  x3  —  a  —  aa;3  i 

reducing,         (4  4-  a)  a;2  =  a  —  2  ;         .  • .     x  =  ±  y  ^777* 


168  APPENDIX. 


MISCELLANEOUS  EXAMPLES   OF  SIMULTANEOUS  EQUATIONS    OF 
THE   SECOND  AND   HIGHER  DEGREES. 


/  y?  _f_  ^3  _  }QXy     t     >     e     n\  , 

1.  Given       •{  }•    to  find  x  and  y. 

<  a;   +  y    =  12         .     ,     .     (2)  i 

Make  a;  =  v  +  w,     and    y  —  v  —  w. 

From  (2),  we  have     (v  +  w)  +  (y  —  70)  =  12 ;         .  • .     v  =  6. 

From  (1),  we  have,     (v  +  w)3  +  (v  —  w)3  =  18  (v2  —  w2)  ; 

or,  reducing,  v3  +  3vw>2  =  9  (vz  —  w2) ; 

substituting  the  value  of  v,  wi  have 

216  +  18u»2  =  9  (36  -  w2),    or    27w2  =  108;        .'.    w=±2, 

hence, 

x  =  v  +  w  =  6±2  =  8    a»l    4       =  v  — w=6^2=4    and    8. 


(z2  +  y-^53     .     .     .      (1)) 

2.  Given  •{  f   to  find  x  and  y. 

(          -  =  14     ...     (2)) 

Multiplying  both  meipW.rs  of  (2)  by  2,  and  adding  and  subtracting, 

we  have 

x2-  +  2a;y  '    «2  =  81  ;         . ' .     x  +  y  =  ±  9, 
xi  _  9/f/f  :c  yZ  _  25  ;         .  • .     x  —  y  =  ±  5  ; 

hence,         «  rz    >   1 ,     and      —  7     y  =  +  2,     and     —  7. 

U«  +  y*  =  82     .     .     .     (1)) 

3.  Given  •{  \  to  find  ar  and  y. 

U  +y  =   4    .    .     .     (2)) 

Raising  both  numbers  of  (2)  to  the  4th  power,  adding  to  (1),  mem 
her  to  member,  and  dividing  by  2, 

x*  -f  2z3y  +  3zV  +  2ary3  +  y*  =  169 ; 


ADDITIONAL   EXAMPLES.  169 

extracting  the  square  root  of  both  members, 

*2  +  xy  +  y2  =  13     .     .     .     (3); 

squaring  both  members  of  (1), 

ar»  +  2xy  +  y2  =  16  .     .     .     (4)  ; 

subtracting  (3)  from  (4),  member  from  member, 

xy  =  3,     or     3xy  —  9    .     .     (5) ; 

subtracting  from  (3),  member  from  member, 

z2  —  2zy  +  y2  =  4 ; 

whence,  x  —  y  =  ±  2    .     .     .    (6). 

Combining  (2)  and  (6), 

x  —  3,     and     1 ;         y  =  1,     and     3. 

( 5*  +  3y   =  19     .     .     (1)  ) 

4.  Given  •<  >   to  find  x  and  y 

('7xz-2yz=lO    .     .     (2)} 

From  (1),  we  find 

_  19  -  3y  2  _  361  -  114y  +  9y2 

~5        ;  ~25~ 

Substituting  in  (2),  and  reducing, 


y 


2_798_       2277 
~  13  ~~         13    ' 


399^      X-2277  ,    159201       399  ±  360 
whence,  y  =  —  ±  ^  — —  +  —_  = ir~ 

759 
hence,  y  =  — ,     and    y  =  3 ; 

by  substitution,       x  = — -,     and    a;  =  2. 

18 


170  APPENDIX. 

(x  +  4y  =  14     .     .     (1)) 

5.  GrtVb.1  •{  >   to  find  x  and  y. 

(y»  +  4a;  =  2y  +  11     (2)) 

From  (1),         „•  =  14  —  4y,     or    4*  =  56  —  16y ; 
subtracting  ana  reducing, 

y2  -  18y  =  -  45  ; 


whence,     y  =  9  ±  ^/  —  45  +  81  =  9  ±  6  =  15     and    3  ; 
hence,  x  =  2     and     —  46. 


z2  +  4y2  =  256  -  4zy     .     .     (1) 

6.  Given        •{  >  to.  find  x  and  y. 

3y2-  a;2  =  39     ...     (2) 


) 

> 
) 


Transposing  in  (1),  and  extracting  the  square  root  of  both  members, 

*  +  2y=±16;         .'.     a;  =±16  —  2y; 
or,  x1  =  256  ^  64y  +  4y2  ; 

substituting  in  (2),  and  reducing, 

y2  =p  64y  =  -  295  ; 


whence,     y  =  ±  32  ±  -y/  -  295  +  1024  =  ±  32  ±  27 ; 

.  • .     y  =  ±    59     and     ±  5  ; 
substituting,  x  =  ±  102     and     ±  6. 

7.  Given  •?  f  to  find  a;  and  y. 

(a;2  +  2:y  =84     .     .     (2)) 

Subtracting  (1)  from  (2),  member  from  member, 
y2  +  xy  =  60     .     .     .     (3)  ; 
adding  (3)  to  (2),  member  to  member, 

x2  +  2xy  +  y2  =  144 ;         . '.     x  -f  y  =  ±  12     .     .     .     (4). 


ADDITIONAL   EXAMPLES.  171 

Dividing  (1)  by  (4),  member  by  member, 

x-y=  ±2; 
hence,  x  —  ±.  7,        y  =  db  5. 

(x-y  =    4     .     .     .     (1)) 

8.  Given  •{  r  to  find  ar 

(       xy  =  45     .     .     .     (2) ) 

From  (1),  x  =  y  +  4,     which,  in  (2),  gives 

y2-f4y  =  45;         .'.     y  =  —  2  ±  -/45  +  4  =  —  2  ±  7 ; 

.'.     y  =  +  5,     and     —  9; 
and  by  substitution,        x  =  +  9,     and     —  4. 

fary  +  *y2  =  12     .     .     .     (1) ) 

9.  Given  \  t   to  find  x  and  y. 

(   x  +  xy3  =  18     .     .     .     (2)  ) 

Dividing  (2)  by  (1),  member  by  member,  and  reducing, 

1-fy3      3  1  —  y  +  yz      3 

,.         .  =  =»     or 


y          ~2' 


g 

by  reduction,  y2  —  -  y  =  —  1  ; 

v 


5          /      ,    ,   25       5       3 

whence,  y  =  -  ±  i  /  —  1  +T^  =  T=tT; 

4V  1644 

1 
.  • .     y  =  2,     and     y  =  — ; 

• 

by  substitution,  a:  =  2,     and    or  =  16. 

1        1 

X    "y    =<i     '      '      (1) 
10.  Given  •{  V  to  find  a;  and  y. 


172  APPENDIX. 

From  (1),  we  find, 

1  1  1  2a       1 

-  =  a »     or    —  =  a2 r*  -5 » 

y  x  y2  as2 

substituting  in  (2),  and  reducing, 

&-a2 

2       ' 

±  i/26  -  a3 


1       a          /6  —  a2   ,    a2       a 

*        —  ~~  —  "4~  \    /  ,  -1-   . ,  — 

'   '  z  ~  2  ~  V      2       r  4  " 


1         a  :p  -1/26  —  a3 
by  substitution,  -  = ^-5 ; 

y  » 

2  ° 

nence,  x  = ,     and    y  = 


a  ±  y  26  —  a2  a  q=  -y/26  —  a2 

(    1-2         4.1-          ft*!  \ 

I  _  _i_  _  =  _     .     .  fl\  J 

11.  Given  J  y2       y  ~  9  \   to  find  *  and  * 

I  -r  _  */  —  <>  fa\  1 

j.   x  —  y  —  A  *  '~4 

Clearing  (1)  of  fractions, 

9z2  +  36a;y  =  85y2     •     •  •     (3). 
From  (2),         x  =  2  +  y ;         .' .    s2  =  4  +  4y  +  y3  ; 
substituting  in  (3),  and  reducing, 

27  9 

y2~io  y  =  io; 


27         /9    ,  729 

wnpnfp  v  — -t-  *  /  —  -4-          •  ~~ 

y  ~  20      V  10  ^  400  ~ 
.  • .    y  =  3     and    y  =  -  — 

17 

by  substitution,          x  =  5    and    a;  =  —  • 


27  db  33 


ADDITIONAL   EXAMPLES.  173 

(f!  +  ^+*  +  *:  =  ?Z..m) 

12.  Given     J  y*      z2      y      *       4         v  '  I  to  find  x  and  y. 
(   *-y  =  2 (2)  J 

X         II 

Make         -  +  -  =  z    (3) ;     whence,  by  squaring, 


hence,  from  (1),  by  substitution  and  reduction, 

35  1          735  ,    1       -  1  ±  6 

*2  +  z  =  T;          .-.     ,=  --±v/T  +  -  =  __; 

5  7 

or,  z  =  -    and     --; 

substituting  the  positive  value  of  z  in  (3),  and  clearing  of  fractions, 

2  _L_      2       .  //I  \ 

From  (2),        *  =  y  +  2 ;         . ' .     z2  =  y2  +  4y  +  4, 
and  zy  =  y2  -f  2y  ; 

substituting  in  (4)  and  reducing, 

hence,  y  =  2     and     y  =  —  4  ; 

by  substitution,         x  =  4     and    y  =  —  2. 

r  xz  +  y2  -f  22  =  84     •     .     •     (1)  >j 
13.  Given  -{    *  +  y  +  z   =  .14    •     •     •     (2)    I     to  find  ar,  y. 

I          22  =  y2  •     •     •     (3)  J  and  2' 

Substituting  in  (1)  and  (2)  the  value  of  y  from  (3),  and  reducing 

z2  4-  2xz  +  z2  =  84  -f  tz     or     a;  +  z  —  t/84  -f  a*     •     •     (4) 
From  (2)  and  (3),  *  + 2  =  14-^7    •     •     (5) 


174:  APPENDIX. 

Equating  the  second  members, 

14  -  ^/xz  —  -v/84  +  xz  -, 
squaring  both  membeis, 

196  -  28  V**  4-  «  =  84  +  xz ; 

.•educing,  -y/arz  =  4,     or    £2  =  16     •     •     •     (6) ; 

hence,  from  (5),  x  +  0  =  10     •     •     •     (7) ; 

1  A 

aobstituting  in  (7)  for  z  its  value  — ,  and  reducing, 

sc 

a2- 10*=  -  16; 


*  =  5  ±  ^/—  16  +  25  =  5  ±  3 ; 
or.  a;  =  8,         a;  =  2 : 

by  substitution,     2  =  2,     2  =  8,     and    y  —  ±  4. 


f  *  +  y  +  -v/#  +  y  =  12  .  •  (1 
14.  Given     {  ;2  +  y  =  41..( 

From  (1),  by  transposition, 

V*+y  =  12  -  (x  +  y)  ; 
squaring  both  members, 

*  +  y  =  144  -  24  (x  +  y)  +  (*  +  J/)2  ; 
reducing,  (a;  +  y)2  —  25  («  +  y)  =  —  144  ; 


25  ~  1^5"  25  ±  7 


whence,  x  -f  y  =  16,     or     x  +  y  =  9. 

Tlie  first  value  does  not  satisfy  (1),  unless  the  radical   have   the 
negative  sign  ;   adopting,  therefore,  the  second   value,  from   which 

x  =  9  —  y,     or     x2  =  81  —  18y  +  y2, 


ADDITIONAL   EXAMPLES. 


which  in  (2)  gives,  after  reduction, 

y2  —  9y  =  —  20 ;     whence, 


T  =  Hr';        .'.    y  =  5    and    y  =  4; 

by  substitution,  x  =  4     and     x  =  5. 

lX3-y3~U1[       .       .       .       (1)1 

15.  Given       •{  f    to  find  x  and  y. 

(x   -y   =      3     ...     (2)) 

Cubing  both  members  of  (2),  subtracting  from  (1),  member  from 
member,  and  dividing  both  members  by  3,  we  have 

xzy  —  -xyz  =  30,     or     (r.  —  y)  xy  =  30     •     .     •     (3)  ; 
dividing  (3)  by  (2),  member  by  member, 

xy  =  W;         .'.     y  =  —; 

X 

substituting  in  (2),  and  reducing, 

xz  -  3*  =  10  ; 

3          I          93? 
whence,  x  =  -  ±  ^  10  +  -  =  -  ±  -; 

.  •  .     x  =  5,         x  =  -  2  ; 
by  substitution,         y  =  2,     and     y  =  —  5. 

(x*  +  r*ffi=    208     •     •     (1)) 

16.  Given    1  [  to  find  a?  and  y, 

-    •     (2)  } 


These  equations  may  be  written, 

•        4  2.  444 

x*+x*y*=    208,     or     xs(xs+ys)=    208     •     .     •     (3), 

^  +  ^=1053,  yVhar^lOSS     ...     (4) 


176  APPENDIX. 

Dividing  (3)  by  (4),  member  by  member, 

aJ   _  208  _  16 
~|  ~~  1053  ~  §1 

y 

extracting  the  4th  root  of  both  members, 


I  '  ' 


substituting  in  (3), 

64         .   16 


=  208;     whence, 


208 

—  y2  =  208,     or    y»  =  729 ;         •'•     y  =  ±  27, 

and  by  substitution,  x  =  ±    8. 

MISCELLANEOUS   PROBLEMS. 

1.  A  courier  starts  from  a  place  and  travels  at  the  rate  of  4  miles 
per  hour ;  a  second  courier  starts  after  him,  an  hour  and  a  half 
later,  and  travels  at  the  rate  of  5  miles  per  hour :  in  how  long  a 
time  will  the  second  overtake  the  first,  and  how  far  will  he  travel  ? 

Let      x     denote  the  number  of  hours  travelled  by  2d  courier  : 
then  will  x  +  1%     "  "         "  "  1st       " 

5x  "  "         "  miles        "  2d       " 

and       4(x  -f  H)    "  "        "      "  "  1st      ;t 

From  the  conditions  of  the  problem, 

bx  =  4  (x  +  li)  ;         .  • .     x  =  6    and     5x  =  30. 

2.  A  person  buys  4  houses  for  $8000 ;  for  the  second  he  gave 
half  as  much  again  as  for  the  first ;  for  the  third,  half  as  much  again 
as  for  the  second  ;  and  for  the  fourth,  as  much  as  for  the  first  and 
third  together  :  what  does  he  give  for  each  1 


ADDITIONAL   EXAMPLES.  177 

Let     x  denote  the  amount  paid  for  1st  house  :  then  will 

ar  + 1         »      «        u          <c      «    2d      " 

a; 
2#  -j —          "       "         "          "      u    3d       u 

3ar  +  ^          "       «         «          «      «    4th      « 
4 

From  the  conditions  of  the  problem, 

Sx  —  8000;  .-.     a:  =1000 

*  +  I  =  1500 

2*  +  ?  =  2250 
4 

3*  +  7  =  3250 
4 

3.  A  and  B  engaged  in  play  :  after  A  had  lost  $20,  he  had  one 
third  as  much  as  B  ;  but  continuing  to  play,  he  won  back  his  $20, 
together  with  $50  more,  and  he  then  found  that  he  had  half  as  much 
again  as  B  :  with  what  sums  did  they  begin  ? 

Let  x  and  y  denote  the  sums  with  which  A  and  B  began. 

Then  from  the  conditions, 

_        _  y +  20 

x  +  50  =  (y  —  50)  X  1 J ; 

whence,        Sx  —    60  =    y  +    20 ;         .  • .     x  =    70 
2x  +  100  =  3y  -  150 ;  y  =  130. 

4.  A  can  do  a  piece  of  wort   in  10  days,  which  A  and  B   ogether 
can  do  in  7  days :  in  how  many  days  can  B  do  it  alone  ? 

Let  x  denote  the  number  of  days. 


178  APPENDIX. 

Since  A  and  B  togethe.   can  do  ^  of  the  •«  ork  in  1  day,  A  can  do 
fa  of  it.  and  B,  -  of  it  in  1  day  ;  hence,  from  the  relations  existing, 

•v 

l 


5.  A  person  has  $650  invested  in  two  parts  :  the  first  part  draws 
interest  at  3  per  cent,  and  the  second  at  3£  per  cent,  and  his  total 
income  is  $20  per  annum  :  how  much  has  he  invested  at  each  rate  ? 

Let  x  denote  the  number  of  dollars  at  3  per  cent :    then  will 
650  —  x  denote  the  number  at  3£. 
From  the  conditions, 

3x   650  -  x 

100  *  100 

whence,        3x  +  2275  —  31  x  =  2000  ; 

.-.  ^  =  275,  or  x  =  550;    .-.  650  -  x  =  100. 
2 

6.  A  boatman  rows  with  the  tide,  in  the  channel,  18  miles  in  1^ 
hours ;  he  rows  near  the  shore  against  the  tide,  which  is  then  only 
three-fifths  as  strong  as  in  the  channel,  18  miles  in  21  hours  :  what 
is  the  velocity  of  the  tide  per  hour  in  the  channel  ? 

Let     x     denote  the  velocity  of  the  tide  in  the  channel : 

Sx 

then,    —          "      "          "  "         near  shore ; 

o 

and       (18  — —  I  -f- 1|  will  denote  the  rate  of  rowing,  neglecting  tide 

(27r\ 
134_       'J-f-oi      «  »  u         « 

hence,  (18  -  '-£]  x  -  -    18  +  -^j  X  - ; 

\  £  I          O  tiv  I  J 

3.c 

or,  12  —  x  =  8  -|-  —  ,         . •.     ar  =  2J. 

o 


ADDITIONAL   EXAMPLES.  179 

7.  A  garrison  had  provisions  for  30  months,  but  at  the  end  of  4 
months  the  number  of  troops  was  doubled,  and  3  months  afterwards 
it  was  reinforced  by  400  troops  more,  and  the  provisions  were  ex- 
hausted in  15  months  :  how  many  troops  were  there  in  the  garrison 
at  first  1 

Let  x  denote  the  number  cf  men  at  first ;  then  will  oQx  denote 
the  number  of  months  that  one  person  could  subsist  on  the  provi- 
sions, or  the  number  of  month'y  rations  in  the  garrison. 

4x     denotes  the  number  of  monthly  rations  used  in  4  months, 
(5x         "  "  "  "  "    the  next  3       " 

(2* +400)8  "  "  "  "  "         "         8       " 

hence,       26*  -f  3200  =  30z,     or     4x  —  3200  ,       .  • .  x  =  800. 

8.  What  is  the  number  whose  square  exceeds  the  number  itself 
by  6?     . 

Let  x  denote  the  number. 

From  the  conditions, 

1  -t  5 

a-2  —  x  —  6  •  •       x  —  1  ±  t/(5  -I-  i  —  • 

x  —  o,          ..      x  —  -j  a:  y  u  -(-  4  _  , 

.  • .     x  =  3     and     —  2. 

9.  Find  two  numbers  such  that  their  sum  shall  be  15,  and  the  sum 
of  their  squares  117. 

Let  x  and  y  denote  the  numbers. 
From  the  conditions  of  the  problem, 

x   +  y    =    15     •     .     .     (1), 
z2  +  y2  =  117     •     •     •     (2). 
From  (1)         x  -  15  —  y,     or    x2  -  225  —  30y  +  y2 ; 


180 


APPENDIX. 


substituting  in  (2)  and  reducing, 

yz  —  I5y  =  —  54, 


15   .      /  .225       15  .   3 


y=—  ±y-54-f  —  =  —  ±-,     or    y  =  9     and     x  --  6, 

or    x  =  9     and     y  =  6. 

10.  A  cask  whose  contents  is  20  gallons,  is  filled  with  brandy  ;  a 
certain  quantity  is  drawn  off  into  another  cask  of  the  same  size, 
after  which  the  latter  is  filled  with  water  :  the  first  cask  is  then  filled 
with  this  mixture  ;  it  then  appears  that  if  6f  gallons  of  this  mixture 
be  drawn  from  the  first  into  the  second  cask,  there  will  be  equal 
quantities  of  brandy  in  each.  How  much  brandy  was  first  drawn 
off? 

Let  x  denote  the  number  of  gallons  first  drawn  off.  Then  will 
20  —  x  denote  the  quantity  remaining  as  well  as  the  quantity  of 

X 

water  added  to  the  second  cask  ;  —•  will  denote   the  quantity  of 
brandy  in  each  gallon  of  the  mixture,  and 

/*•  *y«2 

*X20'     °r     20 

will  denote  the  quantity  of  brandy  returned  to  the  first  cask,  which 
will,  therefore,  contain 


gallons  -of  brandy.     Each  gallon  of  this  new  mixture  will  contain 
^  of  the  brandy  in  the  cask,  or 

400  —  20*  +  x*  m 
400 

nence.  6|  gallons  will  contain 

400  -  20*  +  a* 
60 


ADDITIONAL   EXAMPLES.  181 

gallons ;  and  after  this  is  drawn  off,  10  gallons  must  remain  ;  hemse, 
400  -  20z  +  *2      400  -  20*  +  x* 


20  60 

whence,  800  —  40x  +  2x2  =  600, 

or,  xz  —  20.e  =  100  ; 


=  10; 


.  • .     x  =  10  ±  •/—  100  +  100,     or    x  =  10. 

11.  "What  number  added  to  its  square  will  produce  42? 
Let  x  denote  the  number. 

From  the  conditions  of  the  problem, 
z2  +  x  =  42; 


2 

12.  The  difference  of  two  numbers  is  9,  and  their  sum  multiplied 
by  the  greater  gives  266  :  what  are  the  two  numbers  ? 
Let  x  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 

r M  —  q  /i\ 

x  —  y  —  y  (i), 

z(x  +  y)  =  266     •     (2). 

\          •      **  /  \     / 

From  (1),  y  =  x  —  9  ;         substituting  in  (2). 

x  (2z  —  9)  =  266,     or    x2-  —  -  x  =  133  ; 

m 


9    /     81   9  ±  47 
whence,       x  =  -  ±  \J  133  +  —  = ; 

.-.  x  =  14,    x  =  —  9£; 
whence,  y  =  5,    y  =  —  18^. 

13.  A  person  travelled  105  miles  :  if  he  had  travelled  2  miles 

11 


182  APPENDIX. 

;>er  hour  slower,  he  would  have  been  6  hours  longer  in  completing 
ihe  journey  :  how  many  miles  did  he  travel  per  hour  ? 

Let  x  denote  the  number  of  miles  travelled  per  hour.     Then  will 

105 

denote  the  number  of  hours. 

x 

From  the  conditions, 

105         105 

= +  6,     or     105*  =  105*  -  210  +  6x2  -  12* ; 


x  —  2         x 
reducing,  x2  —  2*  =  35  ; 

.-.     x  =  1  ±  -v/35  +  1  =  1  ±  6  ;         .'.     x  =  7. 

14.  The  continued  product  of  four  consecutive  numbers  is  3024  : 
what  are  the  numbers  ? 

Let  x  denote  the  least  number. 
!  .From  the  conditions  of  the  problem. 

x(x  +  1)  (*  +  2)  (x  +  3)  =  3024, 
or  x*  +  6x3  +  II*2  +  Gx  —  3024  =  0. 

A  superior  limit  of  the  real  positive  roots  is  9  (Art.  279).  Ne- 
glecting the  divisor  1,  and  all  negative  divisors,  we  may  proceed  by 
the  rule  (Art.  285),  as  follows  : 

9,  8,  7,  6,  4,  3,  2, 

-  336,     —  378,      -  432,     —  504,     -  756,      -  1008,      -  1512, 

-  330,     -  372,     -  426,     —  498,     -  750,      -  1002,      -  1506, 

-,       -    83,  -.,       -    334,     -    753, 

.,       .'72,  ..,       _    323,  742, 

••,  ••,  ••,       —    12,  ••,  ••,       —    371, 

ft  0,'K 

•*,  ",  ",  U,  ••,  ••,  OUt», 

•',  ",                 ••,                    1,               ",                    ••,                  ", 

"»  ">                ">                  ">             ">                  "t                " 

Hence,  6  is  the  required  value  of  x,  and  the  numbers  a-  e  6,  7,  8 
and  1). 


ADDITIONAL   EXAMPLES.  183 

15.  Two  couriers  start  at  the  same  instant  for  a  point  39  miles 
distant  ;  the  second  travels  a  quarter  of  a  mile  per  hour  faster  than 
the  first,  and  reaches  the  point  one  hour  ahead  of  him  :  at  what 
rates  do  they  travel  1 

Let  x  denote  the  number  of  miles  per  hour  of  first  courier. 

39 

Then  will  —  denote  the  number  of  hours  he  travels. 
x  . 

From  the  conditions, 


reducing, 


or   *  = 


16.  The  fore-wheels  of  a  wagon  are  5  \  feet,  and  the  hind-wheels 
7£  feet  in  circumference  ;  after  a  certain  journey,  it  is  found  that  the 
fore-wheels  have  made  2000  revolutions  more  than  the  hind-wheels  ^ 
how  far  did  the  wagon  travel  ? 

Let  x  denote  the  number  of  feet. 

From  the  conditions  of  the  problem, 


1197 
multiplying  both  members  by    -5^-, 

BSB 

_  2394000  _  598500 
7W*   ^x     32       8   ' 

57  x  —  42  x  =  598500, 

V 

15  a;  =  598500, 
x=    39900. 


184  APPENDIX. 

17.  A  wine  merchant  has  2  kinds  of  wine ;  the  one  costs  9  shil* 
lings  per  gallon,  and  the  other  5.     He  wishes  to  mix  them  together 
in  such  quantities  that  he  may  have  50  gallons  of  the  mixture,  and 
so  that  each  gallon  of  the  mixture  shall  cost  8  shillings. 

Let  x  and  y  denote  the  number  of  gallons  of  each,  respectively. 
From  the  conditions, 

x+    y  =  50      .     .     .-    .     (1), 
9x  +  5y  =  S(x  +  y)     .     .     (2)  ; 
substituting  for  x  +  y  its  value  in  (2), 

9*  +  5y  =  400    .     .     .     .     (3); 
combining  (1)  and  (3), 

4y  =  50-,         .'.     y  =  12$,     and     x  =  37£. 

18.  A  owes  $1200  and  B,  $2500,  but  neither  has  enough  to  pay 
his  debts.     Says  A  to  B,  "  Lend  me  the  eighth  part  of  your  fortune, 
and  I  can  pay  my  debts."     Says  B  to  A,  "  Lend  me  the  ninth  part 
of  your  fortune,  and  I  can  pay  mine  :"  what  fortune  had  each  1 

Let  x  and  y  denote  the  number  of  dollars  in  the  fortunes  of  A 
and  B. 

From  the  conditions  of  the  problem, 

x  +  |  =  1200,         or         8x  +    y  =    9600, 
o 

y  +  ^  =  2500,         or          x  -f  9y  =  22500 ; 
y 

combining  and  eliminating  x, 

71y  =  170400;         .-.     y  =  2400,        x  =  900. 

19.  A  person  has  two  kinds  of  goods,  8  pounds  of  the  first,  and 
9  of  the  second,  cost  together  $18,46;  20  pounds  of  the  first,  arrd 
16  of  the  second,  cost  together  $36,40:  how  much  does  each  cost 
per  pound  ? 


ADDITIONAL   EXAMPLES.  185 

Let  x  and  y  denote  the  cost  of  a  pound  of  each  in  cents. 
From  the  conditions  of  the  problem, 

Sx+    9y=1846, 
20*  +  16y  =  3640 ; 
combining  and  eliminating  x, 

13y=1950:        .'.    y  =  150,    and    x  =  62. 

20.  What  fraction  is  that  to  the  numerator  of  which  if  1  be 
added  the  result  will  be  £,  but  if  1  be  added  to  the  denominator  the 
result  will  be  J  ? 

Let  x  denote  the  numerator,  and  y  the  denominator. 

From  the  conditions  of  the  problem, 
x+  1       1 


V 

x          1 


or        3x  +  3  =  y, 
or         4x  =  I  +  y  ; 


i+y     4 
hence,  by  combination,    *  =  4    and    y  =  15.     Ans.  y^. 

21.  A  shepherd  was  plundered  by  three  parties  of  soldiers.  The 
first  party  took  -J-  of  his  flock  and  £  of  a  sheep ;  the  second  took  £ 
of  what  remained  and  ^  of  a  sheep ;  the  third  took  -J  of  what  then 
remained  and  ^  of  a  sheep,  which  left  him  but  25  sheep :  how  many 
had  he  at  first  ? 

Let  x  denote  the  number  of  sheep.  Then,  after  being  plundered 
by  the  1st  party,  he  would  have 

O_  1 

sheep ; 


4 

after  being  plundered  by  the  2d  party,  he  would  have 
3x  —  I       /3x  —  1    .    1\       x  —  I 


1       / 
\ 


12 


186 


APPENDIX. 


after  being  plundered  by  the  3d  party,  he  would  have 
x  —  1        /x  —  1       l\       Z  —  3 


1\       x 
27  = 


2 
from  the  conditions  of  the  problem, 

-  =  25,     or     x  —  3  =  100;         .-.     x  —  103. 

22.  What  two  numbers  are  those  whose  product  is  63,  and  the 
square  of  whose  sum  is  equal  to  64  times  the  square  of  their  dif 
ference  1 

Let  x  and  y  denote  the  two  numbers. 
From  the  conditions  of  the  problem, 

xy  -  63 (1), 

(x  +  yY  =  64  (x  -  y)2     .     .     (2); 
extracting  the  square  root  of  both  members  of  (2), 

x  -{-  y  =  8  (x  —  ?/),     or     7#  =  9y  ;          . ' .     x  =  2-y  ; 
substituting  in  (1),  f  y2  =  63  ; 

.  • .     y2  —  49     and     y  =  7,     also     x  =  9. 

23.  The  sum  of  two  numbers  multiplied  by  the  greater  gives 
209  ;  their  sum  multiplied  by  their  difference  gives  57  :  what  are 
the  two  numbers  ? 

Let  x  and  y  denote  the  numbers. 
From  the  conditions  of  the  problem, 

(x  +  y)  x  =  209,     or    x2  4  zy=  209    .     .     (1), 
(z  +  y)(x-y)=    57,     or     a*  -  y*  =    57     .     .     (2); 
subtracting  (2)  from  (1),  member  from  member, 

adding  (3)  and  (1),  member  to  member, 

x2  +  2xy  4-  v2  =•  361 ;         .  • .     x  -4-  y  =  ]  9  ; 


ADDITIONAL   EXAMPLES.  187 

209 
hence,  from  (1),  £  =  -—-  =  11;     also,     y  =  8. 

24.  Three  numbers  are  in  arithmetical  progression  ;  their  sura  is 
15,  and  the  sum  of  their  cubes  is  495  :  what  are  the  numbers  1 
Let  x,  y  and  z  denote  the  numbers, 
From  the  conditions  of  the  problem, 

y  —  x  =  z  —  y  .  .  (1) 
x  -hy  +z  =  15  .  .  (2) 
x3  +  y3  +  z3  =  495  .  .  (3)  ; 

from  (1),         2y  =  z  +  or,       which  in  (2),  gives       y  =  5 ; 
substituting  in  (2)  and  (3), 

z   +   x  =    10     .     .     (4) 
03  -f  x3  =  370     .     .     (5)  ; 
dividing  (5)  by  (4),  member  by  member, 

22  -  zx  '+  x*  =  37     .     .     (6)  ; 
squaring  both  members  of  (4), 

z2  +  2zx  +  x2-  =  100     .     .     (7)  ; 

combining  (6)  and  (7), 

21 
'6zx  =  63,     or  *  zx  =  21 ;         .  * .      z  =  —  ; 

X 

substituting  in  (4), 

21 

*  +  —  =  10,       or       xz  —  10*  =  -  21 ; 


.-.     x  =  5  ±  -/—  21  +  25  =  5  ±  2  ;         hence,     x  =  7.     or  3. 

25.  Divide  the  number  16  into  two  parts  such  that  25  times  the 

square  of  the  first  shall   be  equal  to  9  times   the  square  of  the 
second. 


188  APPENDIX. 

Let  x  denote  one  part ;  then  will    16  —  x   denote  the  other. 
From  the  conditions, 

25*2  =  9  (256  -  32*  +  *2)  =  2304  -  288*  +  9*2 ;     . 
reducing,  xz  +  18*  =  144  ; 


.-.     x  =  —  9  ±  -/144  +  81  =  —  9  ±  15,     or    *  =  6, 

since  the  negative  value  does  not  satisfy  the  problem  understood  in 
the  numerical  sense. 

26.  There  are  two  numbers  such  that  the  greater  multiplied  bj> 
the  square  root  of  the  less  is  18,  and  the  less  multiplied  by  the 
square  root  of  the  greater  is  12  :  what  are  the  numbers  ? 

Let  *  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 

y^=lS     .     .     (1) 
*V7=12     .     .     (2); 
multiplying  (1)  by  (2),  member  by  member, 

y^r)3"=216;         .'.     xy  -  36     .     .     (3); 
adding  (1)  and   (2),  member  to  member, 

(V^+  V7)  V*y  =  30     .     .     (4), 
or  -\/x  +  yV  =  5  ; 

squaring  both  members, 

*  +  y  +  2V*7=25     .     .     (5), 
or  *  +  y  =  13  (6); 

combining  (3)  and  (6), 

x  =  9.         y  =  4. 


ADDITIONAL   EXAMPLES.  189 

27.  What  two  numbers  are  those  the  square  of  the  greater  of 
which   being   multiplied   by  the  lesser  gives  147,  and  the  square 
of  the  lesser  being  multiplied  by  the  greater  gives  63  1 

Let  x  and  y  denote  the  numbers. 
From  the  conditions  of  the  problem, 

*2y  =  147     •     •     (1) 
xy*=    63     .     .     (2); 
multiplying  (1)  and  (2),  member  by  member, 

*V  =  9261     .     .     (3), 

or  xy=      21     .     .     (4); 

dividing  (2)  by  (4),  member  by  member, 

y  =  3  ;         in  like  manner,         x  =  7. 

This  method  of  solution  might  be  applied  to  the  equations  of  the 

preceding  example. 

•  . 

28.  There  are  two  numbers  whose  difference  is  2,  and  the  product 
of  their  cubes  is  42875  :  what  are  the  numbers  ? 

Let  x  and  y  denote  the  numbers. 

From  the  conditions  of  the  problem, 

*-y  =  2     .     .     (1) 
X3y3  _  42875     .     .     (2) ; 
extracting  the  cube  root  of  both  members  of  (2), 

35 

ary  =  35;         .-.     y  =  — ; 

substituting  and  reducing, 

**  -  2*  =  35, 

x   —  1  ±  -/S5  +  1  =  1  ±  6; 
.  • .     x  =  7,     and     —  5,     y  =  5,     and     —  7. 


190  APPENDIX. 

29.  A  sets  out  from  C  towards  D,  and  travels  8  miles  each  day ; 
after  he  had  gone  27  miles,  B  sets  out  from  D  towards  C,  and  goes 
each  day  ^  of  the  whole  distance  from  D  to  C ;  after  he  had 
travelled  as  many  days  as  he  goes  miles  in  each  day,  he  met  A  • 
what  is  the  distance  from  D  to  C? 

Let  x  de  lote  the  number  of  miles  from  D  to  C. 

X 

Then,     —     will  denote  the  number  of  miles  B  travels  per  day, 
£\j 

also  the  number  of  days  that  he  travels  ; 

a;2 
hence,     --^-r       denotes  the  number  of  miles  travelled  by  B, 

27  +  Sx       "        "         "  "  "  A. 

"From  the  conditions  of  the  problem, 

*2  _L  07  _L  Sx  _ 
lOO4"         "^20" 

clearing'of  fractions  and  reducing, 

x2-  -  240*  =  -  10800  ; 


•    .-.     x  -  120  ±  -/-  10800  +  14400  =  120  ±  60 ; 
whence,  x  =  60,         x  =  180. 

30.  There  are  three  numbers  ;  the  difference  of  the  differences  of 
the  1st  and  2d,  and  2d  and  3d,  is  4  ;  their  sum  is  40,  and  their  con 
tinued  product  is  1764  :  what  are  the  numbers  ? 

Let  #,  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

(*-y)-(y-*)=       4    .    .     (l) 
x  +  y  +  z=      40     .     .     (2) 
.     .     (3); 


ADDITIONAL   EXAMPLES.  191 

combining  (1)  and  (2),  eliminating  x  and  2, 

3y  =  36;         .-.     y  =  12; 
substituting  in  (2)  and  (3), 

x  +  z=    28     .     .     (4) 

xz  =  147     .     .     (5); 
combining  (4)  and  (5), 

x  =  7,     or     21  ;         y  =  21,     or     7. 

31.  There  are  three  numbers  in  arithmetical  progression  :  the 
sum  of  their  squares  is  93,  and  if  the  first  be  multiplied  by  3,  the 
second  by  4,  and  the  third  by  5,  the  sum  of  the  products  will  be  66 : 
what  are  the  numbers  ? 

Let  x  denote  the  first  number,  and  y  their  common  difference. 

From  the  conditions  of  the  problem, 

*2  +  (*  +  2/)2  +  (*  +  2y)2  =  93     .     .     (1) 
3*  +  4  (x  +  y)  +  5  (x  +  2y)  =  66     .     .     (2)  ; 

performing  indicated  operations  and  reducing, 

3z2  -f  5y2  +  Gxy  =  93     .     .     (3) 
12*  -f  I4y  =  66,     or     6*  +  7y  =  33     .     .     (4). 

33  —  Qx 
From  (4),  y  = ; 

1089  —  396*  -f-  36z2  33*  —  6*2 

.'.     y*--  -&-        «i         and        *y=     —^—., 

substituting  in  (3)  and  reducing, 

^  _  198  290 

"  "25^  *  ~  "  "25  ' 


192  APPENDIX. 


-„  ,       ,      296   ,   9801       99  ±  49 
whence,          x  =  -  ±x/ -  —  +  —  = —_; 

148 


Taking  the  second  value  of  x,  we  find    y  =  3,     and  the  numbers 
are  2,  5  and  8. 

The  problem  supposes  the  numbers  entire,  therefore  the  1st  value 
of  x  is  not  used. 

32.  There  are  three  numbers  in  arithmetical  progression  whose 
sum  is  9,  and  the  sum  of  their  fourth  powers  is  353  ;  what  are  the 
numbers  ? 

Let  a;,  y  and  z  denote  the  numbers. 
From  the  conditions  of  the  problem, 

2y  =  x  +  z     .     .     (1) 
x  +  y  +  z  =  9     .     .     (2) 
z*  +  y*  4-  s4  =  353    .     .     (3). 
From  (1)  and  (2)  we  find  y  =  3; 

substituting  in  (2)  and  (3), 

x   +  z   =      6     .     .     (4) 
z*  +  z*  =  272     .     .     (5)  ; 
raising  both  members  of  (4)  to  the  4th  power, 

a;4  +  4X3Z  _|_  QXZZZ  _j_  4XZ3  +  s*  _  J296     .     .     (6) ; 

adding  equations  (5)  and  (6),  member  to  member,  and  dividing  by  2 

Z*  +  2*3z  4-  3z222  +  2z23  -f  2*  =  784    .     .     (7); 
extracting  the  square  root  of  both  members, 

22  -f  xz  +  z2  =  28     .     .     (8); 


ADDITIONAL   EXAMPLES.  193 

squaring  both  members  of  (4), 

z2  +  2xz  +  02  =  36    .     .     (9); 

from  (8)  and  (9)  we  find 

xz  =  8     .     .     (10); 

from  (4)  and  OO)  we  get 

x  =  2,     or    4 ;  s  =  4,     or    2 : 

hence,  the  numbers  are  2,  3  and  4. 

33.  How  many  terms  of  the  arithmetical  progression  1,  3,  5,  7, 
&c.,  must  be  added  together  to  produce  the  6th  power  of  12  ? 
The  6th  power  of  12  is  2985984. 
From  Art.  175  we  have  the  formula, 


d-2a± 

w=  — 


Ir.  the  present  case,        o  =  1,    d  =  2,    and     S  =  2985984  ; 


-V/16  X  2985984       ,WM 
substituting,         n  =  — — =  1728. 

34.  The  sum  of  6  numbers  in  arithmetical  progression  is  48  ;  the 
product  of  the  common  difference  by  the  least  term  is  equal  to  the 
number  of  terms  :  what  are  the  terms  of  the  progression  ? 

Let  x  denote  the  1st  term,  and  y  the  common  difference. 

From  the  conditions  of  the  problem, 

Qx+  15y  =  48,    *y  =  6;        .   .    y=^; 

substituting  and  reducing, 

a*  -  8*  =  -  15 ; 

.-.     x  =  4  ±  •/—  15  +  16  =  4  ±  1,    or    *  =  5,    *  =  3; 
whence,  y  =  f  >        y  =  2 : 


194:  APPENDIX. 

hence,  the  series  is       3.5.7.9.11.13, 
or  5  .  6£  .  7f  .  8£  .  9f  .  11. 

35.  What  is  the  sum  of  10  square  numbers  whose  square  roots 
are  in  arithmetical  progression  the  least  term  of  which  is  3,  and  the 
common  difference  2? 

Let  x  denote  the  sum. 

The  progression  of  roots  is 

3.5.7.9.11.13.15.17.19.21, 
and  the  series  of  squares, 

9 .  25  .  49  .  81  .  121  .  169  .  225  .  289  .  361  .  441. 
1st  order  of  diffs,  16,  24,  32,  40,  &c., 

2d  order  of  diffs,  8,  8,  8,  &c., 

3d  order  of  diffs,  0,  0,  &c. 

From  Art.  210,  making 

S'  =  x,     a  =  9,     n  =  10,     ^  =  16,     dz  =  8,     d3  =  0,     &c. 
x  =  90  +  45  x  16  +  120  x  8  --=  1770. 

36.  Three  numbers  are  in  geometrical  progression  whose  sum  i« 
95,  and  the  sum  of  their  squares  is  3325:  what  are  the  numbers? 

Let  a;,  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

y*  =  a*     .     .     (1) 
xz  +  y»  +  z2  =  3325     .     .     (2) 

x   +  y   +  z   =      95     .     .     (3) ; 
combining  (1)  and  (2), 

x*  +  2xz  \-  22  =  3325  +  xz    ,     .     (4)  ; 


ADDITIONAL   EXAMPLES.  19 

combining  (1)  and  (3), 


x  -f  v^xz  +  z  =  95     .     .     (5)  ; 
from  (4)  and  (5), 

x  +  z  =  ^3325  +  xz 

x  +  2  =  95  — 


hence,  -y/3325  +  xz  =  95 

squaring  both  members, 

3325  +  xz  =  9025  —  190  </xz  +  xz  ; 

.-.     T/XZ  =  30,     or     xz  =  900     .     .     (6)  ; 
substituting  in  (5),  x  -j-  z  =    65     .     .     (7)  • 

from  (6)  and  (7),         *  =  20     and    45, 
y  —  45     and     20. 

37.  Three  numbers  are  in  geometrical  progression  :  the  difference 
of  the  first  and  second  is  6  ;  that  of  the  second  and  third  is  15.: 
what  are  the  numbers  ? 

Let  a;,  y  and  z  denote  the  numbers. 

From  the  conditions  of  the  problem, 

y*  —  xz     .     .     (1) 
x  —  y  =  —    6;        .'.     x  —  y  —    6 
y  —  z—  —  15;         .'.     z=y+15, 
and  xz  —  yz  +  9y  —  90; 

substituting  in  (1)  we  find        y  =  10  ; 

.  •  .     x  =  4,     and     z  •=.  25. 

38.  There  are  three  numbers  in  geometrical  progression  ;  the  sum 


196  APPENDIX. 

of  the  first  and  second  is  14,  and  the  difference  of  the  second  and 
third  is  15  :  what  are  the  numbers  ? 
Let  x,  y  and  z  denote  the  numbers. 
From  the  conditions  of  the  problem, 
y*  =  xz     .     .     (1) 
«  +  y  =  14;         .'.     37=14  —  y 
z  —  y  =  15;         .-.     z  =  15  -f  y, 
and  xz  —  210  —  y  —  yz; 

ft 

substituting  in  (1),  y2  +  ^  =  105 ; 


1        -  1  ±  41  21 

***iiffi« — 4— =10'  and  -«•> 

taking  the  1st  value  of  y,  we  find 

x  =  4,         s  =  25. 

39.  A,  B  and  C  purchase  coffee,  sugar  and  tea  at  the  same  prices ; 
A  pays  $11,62-|  for  7£  pounds  of  coffee,  3  pounds  of  sugar,  and  2£ 
pounds  of  tea  ;  B  pays  $16,25  for  9  pounds  of  coffee,  7  pounds  of 
sugar,  and  3  pounds  of  tea;  C  pays  $12,25  for  2  pounds  of  coffee, 
5£  pounds  of  sugar,  and  4  pounds  of  tea :  what  is  the  price  of  a 
pound  of  each  1 

Let  x,  y  and  z  denote  the  number  of  cents  that  the  coffee,  sugar 
and  tea  cost,  respectively. 

From  the  conditions  of  the  problem, 

^+    fy  +  21z  =  1162$  .  .  (1) 

9*+    7y  +  3z=  1625  .  •  (2) 

2  x  +  5£y  -f  4  z  =  1225  .  .  (3)  ; 


ADDITIONAL   EXAMPLES.  197 

clearing  (1)  and  (3)  of  fractions, 

30*  +  12y  +  90    =  4650       •     •     (4) 
4  ar+lly-f  8z    -2450       •     •     (5). 

From  (2)  and  (4), 

3*  —  9y  =  —  225,    or    x  —  3y  =  —  75     •     •     (6) ; 

from  (2)  and  (5), 

QQx  +  23y  =  5650     .     •     (7); 

from  (6)  and  (7),  y  =  50; 

by  substitution,  x  =  75,  z  =  200. 

40.  Divide  100  into  2  such  parts  that  the  sum  of  their  square 
roots  shall  be  14. 

Let  x  denote  the  first  part. 

From  the  conditions  of  the  problem, 

•y/ar-f  -v/100  —  x  =  14; 
squaring  both  members  and  reducing, 


•V/l  00*  -  a?  =  48 ; 
squaring  both  members  and  reducing, 

*2-100r=  -2304; 


.-.     x  =  50  ±  -/-  2304  -I-  2500  =  50  ±  14, 
x  =  64,     and     36. 

41.  In  a  certain  company  there  were  three  times  as  many  gentle- 
men as  ladies ;  but  afterwards  8  gentlemen  with  their  wives  went 
away,  and  there  then  remained  five  times  as  many  gentlemen  as 
ladies  :  how  many  gentlemen,  and  how  many  ladies  were  there 

originally  ? 

12 


198  APPENDIX. 

Let  3z  denote  the  number  of  gentlemen ;  then  will  x  denote  the 
number  of  ladies. 

From  the  conditions  of  the  problem, 

3x  —  8  =  5  (x  —  8) ; 
.  • .     x  =  1 6,     and     3x  =  48. 

42.  Find  two  quantities  such  that  their  sum,  product,  and  the 
difference  of  their  squares,  shall  all  be  equal  to  each  other. 

Let  x  and  y  denote  the  quantities. 
From  the  conditions  of  the  problem. 

x  +y   -xy     .     .     (1) 

xz  —  y2-  —  xy     •     •     (2)  ; 
by  division  of  (2)  by  (1),  we  have 

x  —  y  =  1,     or     a;  =  y  -f-  1  ; 
substituting  in  (1), 

2y  +  1  =  yz  +  y,    -or     y2  —  y  =  1 ; 

.  1        /T~T 

whence, 

hence,  x  = 

43.  A  bought  120  pounds  of  pepper,  and  as  many  pounds  of 
ginger,  and  had  one  pound  of  ginger  more  for  a  dollar  than  of 
pepper ;  the  whole  price  of  the  pepper  exceeded  that  of  the  gingei 
by  6  dollars :   how  many  pounds  of  pepper,  and   how  many  of 
ginger  had  he  for  a  dollar  ? 

Let  x  denote  the  number  of  pounds  of  pepper  for  a  dollar. 


ADDITIONAL   EXAMPLES.  199 

From  the  conditions  of  the  problem, 


The  negative  value  does  not  conform  to  the  conditions  of  the 
special  problem. 

44.  Divide  the  number  36  into  3  such  parts  that  the  second  shall 
exceed  the  first  by  4,  and  that  the  sum  of  their  squares  shall  be 
equal  to  464. 

Let  #,  y  and  z  denote  the  parts. 
From  the  conditions  of  the  problem, 

x  +  y  +  z  =  36     .     •     (1) 
y-x  =  4       .     .     (2) 
xi  +  y2-  +  z2  =  464  •     •     (3)  ; 

from  (1),  z2  +  2xy  +  y2  =  1296  -  T2z  +  z2     •  '  •     (4)  ; 

from  (2),  z2  —  2xy  +  y2  =  16        ......     (5)  ; 

adding  (4)  and  (5),  member  to  member. 

2*2  +  2y2  =  1312  —  72z  +  z2     •     •     (6)  ; 
from  (3),          2*2  +  2y2  =    928  -    2z2  •     .     .     .     (7)  ; 
equating  the  second  members  and  reducing, 
22  -  24*  =  -  129  ; 


.-.     z  =  12  ±  -V/-128  +  144  =  12  ±  4; 
hence,  z  —  16,         z  =  8 ; 

substituting  the  first  value  in  (1), 

*  +  y  =  20     •     •     (8); 


200  APPENDIX. 

from  (2)  and  (8),  y  =  12    and    x  =  8. 

.45.  A  gentleman  divided  a  sum  of  money  among  4  persons,  so 
that  what  the  first  received  was  ^  that  received  by  the  other  three , 
what  the  second  received  was  £  that  received  by  the  other  three ; 
what  the  third  received  was  -j  that  received*  by  the  other  three,  and 
it  was  found  that  the  share  of  the  first  exceeded  that  of  the  last  by 
$14 :  what  did  each  receive,  and  what  was  the  whole  sum  divided  ? 
Let  x,  y,  z  and  w  denote  the  number  of  dollars  that  each  received. 
From  the  conditions  of  the  problem, 

2#  =  y  +  z  -f  w  '  •  (1) 
3y  =  x  +  s  +  w  •  •  (2) 
42  =  x  +  y  +  w  -  •  (3) 

x  —  w  =  14     •     •     (4) ; 
from  (2)  and  (3), 

x  +  w  —  3y  —  z 

x  -f-  w  =  4z  —  y  ;     whence,     3y  —  z  =.  4z  —  y, 
01  4y  =  5s,         z  —  A  y     .     .     (5) ; 

from  (4),  w  =  x  —  14    •    •     (6) ; 

substituting  the  values  of  w  and  z  in  (1)  and  (2), 
•2x  =  y  +  f  y  +  x  —  14 

3y  —  x  -\-  %y  -\-  x  —  14;     whence,  by  reduction, 
5*  —    9y  =  —  70 
10*-lly=       70; 
,  *.     x  =  40,     y  =  30 ;     and  by  substitution,    z  =  24,     to  =  26. 

.  46.  A  woman  bought  a  certain  number  of  eggs  at  2  for  a  penny, 
and  as  many  more  at  3  for  a  penny,  but  on  selling  them  at  the  rate 


ADDITIONAL  EXAMPLES.  201 

of  5  for  2  pence,  she  lost  4  pence  by  the  bargain  ;  how  many  did 
she  buy  ? 

X  £ 

Let  x  denote  the  number  at  each  price.     Then  will     „  +  5- 

i        o 

denote  the  number  of  pence  paid,  and     — — - — -      will  denote  the 

0 

number  of  pence  received. 

From  the  conditions  of  the  problem, 

x       x       %(x  +  x\ 

^  +  3  =    V  5      '  +  4  ;         reducing,         x  =  120. 

47.  Two  travellers  set  out  together  and  travel  in  the  same  direc- 
tion ;  the  first  goes  28  miles  the  first  day,  26  the  second  day,  24  the 
third  day,  and  so  on,  travelling  2  miles  less  each  day  ;  the  second 
travels  uniformly  at  the  rate  of  20  miles  a  day  :  in  how  many  days 
will  they  be  together  again  1 

Let   x   denote   the    required   number  of  days.      The    distance 
travelled  by  the  first  in  x  days  is 

[(Art.  176),  since    a  =  28,     d  =  —  2,     and    n  =  a;],     denoted  by 
\x  [56  -  (x  —  I)2],     or    29*  —  a:2; 

and  the  distance  travelled  by  the  second  is  denoted  by  20x: 
hence,  we  hare 

29a;  —  xz  =  20ar,     or    x  =  9. 

48.  A  farmer  sold  to  one  man  30  bushels  of  wheat  and  40  of 
barley,  for  which  he  received  270  shillings.     To  a  second  man  he 
gold  50  bushels  of  wheat  and  30  of  barley,  at  the  same  prices,  and 
received  for  them  340  shillings  :  what  was  the  price  of  each  ? 

Let  x  denote  the  number  of  shillings  for  1  bushel  of  wheat, 
and      y       "         "         "•  "  "         "  barley. 


202  APPENDIX. 

From  the  conditions  of  the  problem, 

30*  +  40y  =  270     •  •     (1) 

50a;  +  30y  =  340     •  .     (2)  ; 
whence,                 HOy  =  330,     or    y  =  3;      hence,     x  =  5. 

49.  There  are  two  numbers  whose  difference  is  15,  and  half  their 
product  is  equal  to  the  cube  of  the  lesser  number  :  what  are  the 
numbers  ? 

Let  x  and  y  denote  the  numbers  ; 
from  the  conditions  of  the  problem, 

x  —  y  =  15 

xy  =  2y3  •  or,  x  =  2y2  ; 

substituting  and  reducing, 


--  v  =  —  : 
2y       2  ' 


1  5        1         1±11 


5  25 

hence,        y  =  3,     and     —  -  ;     also,     x  =  18,     and     —  • 


50.  A  merchant  has  two  barrels  and  a  certain  number  of  gallons 
of  wine  in  each.  In  order  to  have  an  equal  quantity  in  each,  he 
drew  as  much  out  of  the  first  cask  into  the  second  as  it  already 
contained  ;  then  again  he  drew  as  much  out  of  the  second  into  the 
first  as  it  then  contained  :  and  lastly,  he  drew  again  as  much  from 
the  first  into  the  second  as  it  then  contained,  when  he  found  that 
there  was  16  gallons  in  each  cask  :  how  many  gallons  did  each 
originally  contain  ? 


ADDITIONAL    EXAMPLES.  203 

Let  x  denote  the  n  imber  of  gallons  in  the  first  cask,  and  y  the 
number  in  the  second  ; 

x  —  y  will  denote  the  quantity  in  the  first  cask  after  the  first  drawing, 
and  2y  the  quantity  in  the  second  cask ;  after  the  second  drawing, 
2y  —  (x  —  y)  or  3y  —  x  will  denote  the  quantity  in  the  second, 
and  2«  —  2y  the  quantity  in  the  first  cask ;  after  the  third  drawing, 
2or  —  2y  —  (3y  —  x}  or  3x  —  5y  will  denote  the  quantity  in  th« 
first  cask,  and  Qy  —  2x  the  quantity  in  the  second. 

From  the  conditions  of  the  problem, 

3*  —  5y  =  16 

6y  —  2z  =  16. 
By  combination, 


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